Hi, This is Anitha@ Satyanarayana Reddy, Welcome To My Channel AS Tutorial తెలుగులో.
In ASTutorialతెలుగులో I will share with you the concepts of Maths in inter.
AS Tutorialతెలుగులో Provides Concepts and Problems for Inter Maths Students In Telugu.
హాయ్, ఇది అనిత @ సత్యనారాయణ రెడ్డి, వెల్ కమ్ మై ఛానల్ AS Tutorial తెలుగులో,
AS Tutorial తెలుగులో ఇంటర్ లో మ్యాథ్స్ భావనలను మీతో పంచుకుంటాను.
AS Tutorial తెలుగులో ఇంటర్ మ్యాథ్స్ విద్యార్థులకు తెలుగులో కాన్సెప్ట్స్ అండ్ ప్రాబ్లమ్స్ ను అందిస్తోంది.
My Youtube Channel: https://www.youtube.com/channel/UCtL8sDWcUlJX_7Npr9460lA
Simple Equations | LHS to RHS and RHS to LHS, RULE ||AS Tutorial తెలుగులో || 2021|| : https://youtu.be/2Vkb_jYx_YM
Trigonometry// ALL-SILVER-TEA-CUPS Rule ||ASTutorialతెలుగులో || 2021|| :https://youtu.be/UqwfCBsJVMY
‘Locus’ Concept and Problems For Inter First Year Maths 1B||AS Tutorial తెలుగులో || 2021||(PART – 1):https://youtu.be/RAtxfld4ywQ
‘Locus’ Concept and Problems For Inter First Year Maths 1B||AS Tutorial తెలుగులో || 2021||(PART – 2): https://youtu.be/E6IaE09YsbM
Trigonometry Introduction, Sides of Right Triangle and Trigonometry Ratios |ASTutorialతెలుగులో 2021:https://youtu.be/Elt7Uy_yfLU
Trigonometry || How To Prove Specific Angles In Trigonometry| || AS Tutorialతెలుగులో|| || 2021|: https://youtu.be/Fh4Vc66swQQ
Trigonometry| How to remember Specific Angles Table in Trigonometry |ASTutorialతెలుగులో || 2021||: https://youtu.be/D3CZ1ijq7OM
Trigonometry| Trigonometric Identities In Trigonometry|ASTutorialతెలుగులో || 2021|| :https://youtu.be/r1NZyagpN90
‘Transformation Of Axes’ Concept and Problems For Inter Maths 1B||AS Tutorial తెలుగులో || 2021|| P 1: https://youtu.be/9qJJuZzKtJ8
‘Transformation Of Axes’ Concept and Problems For Inter Maths 1B||AS Tutorial తెలుగులో || 2021|| P 2 :https://youtu.be/lAfuMQ1YU3c
Playing With Numbers||Divisibility Rules In Telugu ||AS Tutorial తెలుగులో || || 2021|| : https://youtu.be/8I1izylDH_o
Playing With Numbers||Even| Odd|Prime|Composite|Co-Prime &Twin- Prime Numbers|| AS Tutorial తెలుగులో : https://youtu.be/TZeleb6vd1I
Logarithms|| Product Rule| Quotient Rule| and Power Rule||AS Tutorial తెలుగులో || || 2021||: https://youtu.be/ssYCXUc0Hf8
Periodic Functions| Trigonometric Ratios Up To Transformations ||AS Tutorial తెలుగులో || || 2021|| : https://youtu.be/aeOEDgQpVu0
Please Do Not Issue A” Copyright Strikes” Against The Channel As It Affects My Channel And All Previous Work.
If I Uploaded Videos Or Music That Is Yours And You Want It Removed.
Then Please Just Massage Me And I Will Remove The Whole Video In Fewer 12Hours.
*Thanks For Watching This Video*
The post AS Tutorial For Math Classes / YouTube first appeared on Basics In Maths.]]>Telangana BIE Maths Reduced Syllabus(2021) very useful I.P.E exam.
Inter 1st year Maths Reduced Syllabus | Click Here |
Inter 2nd year Maths Reduced Syllabus | Click Here |
The post Telangana BIE Maths Reduced Syllabus(2021) first appeared on Basics In Maths.]]>
Errors and Approximations
Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions
Sol:
Given y = f(x) = x^{2} + x at x = 10, ∆x = 0.1
∆y = f (x + ∆x) – f (x)
= f (10 + 0.1) – f (10)
= f (10.1) – f (10)
= (10. 1)^{2} + 10.1 – (10^{2} + 10)
= 102.01 + 10.1 – 100 – 10
= 112.11 – 110
= 2.11
dy = f’ (x) ∆x
= (2x + 1) (0.1)
= [2(10) + 1] (0.1)
= 21 × 0.1
= 2.1
Sol:
Given y = cos x, x = 60^{0} and ∆x = 1^{0}
∆y = f (x + ∆x) – f (x)
= cos (60^{0} + 1^{0}) – cos 60^{0}
= cos (61^{0}) – 0.5
= 0.4848 – 0.5
= – 0.0152
dy = f’ (x) ∆x
= – sin x (1^{0})
= – sin 60^{0} × 0.0174
=– 0.8660 × 0.0174
= – 0.0150
Sol:
y = x^{2} + 3x + 6
∆y = f (x + ∆x) – f (x)
= f (10 + 0.01) – f (10)
= f (10.01) – f (10)
= (10.01)^{2} + 3 (10.01) + 6 – (10^{2} + 3 (10) + 6)
= 100. 2001 + 30.03 + 6 – 100 – 30 – 6
=130. 2301 – 130
= 0.2301
dy = f’ (x) ∆x
= (2x + 3 + 0) (0.01)
= (2× 10 + 3) (0.01)
= 23 × 0.01
= 0.23
Sol:
The side of a square is increased from 3cm to 3.01cm find the approximate increase in the area of the square.
Sol:
Let x be the side of the square and area be A
Area of the square A = x^{2}
x = 3 and ∆x = 0.01
∆A = 2x × ∆x
= 2(3) (0.01)
= 6 × 0.01
= 0.06
If an increase in the side of a square is 2% then find the approximate percentage of increase in its area.
Sol:
Let x be the side of the square and A be its area
A = x^{2}
∆A = 2x × ∆x
The approximate percentage error in area A
= 2 × 2 =4
From the following. Find the approximations
Sol:
Let f(x) = , where x = 1000 and ∆x =– 1
Approximate value is
f (x + ∆x) = f(x) + f’ (x) ∆x
= 10 – 0. 0033
= 9.9967
Sol:
Sol:
(iv) Sin 62^{0}
Sol:
Let f(x) = sin x, where x = 60^{0} and ∆x =2^{0}
Approximate value is
f (x + ∆x) = f(x) + f’ (x) ∆x
= sin 60^{0} + cos x (2^{0})
= sin 60^{0}+ cos 60^{0} (0.0348)
= 0.8660 + 0.5 × 0.0348
= 0.8660 + 0.0174
=0.8834
The radius of a sphere is measured as 14cm. Later it was found that there is an error of 0.02cm in measuring the radius. Find the approximate error in the surface area of the sphere.
Sol:
Given r = 14 cm and ∆r =0.02cm
Surface area of sphere =A = 4π r^{2}
∆A = 8π r ∆r
= 8 ×3.14× 14 × 0.02
= 7.0336
The post Errors and Approximations || V.S.A.Q’S|| first appeared on Basics In Maths.]]>
Find f’ (x) for the following functions
(i) f(x) = (ax + b) (x > -b/a)
Sol:
Given f(x) = (ax + b)^{ n}
f’ (x) = n (ax + b)^{ n – 1 } (ax + b)
= n (ax + b) ^{n – 1 }a
= an (ax + b) ^{n –} 1
(ii) f(x) = x^{2} 2^{x} log x
Sol:
Given f(x) = x^{2} 2^{x} log x
f’ (x) = ^{ }(x^{2}) 2^{x} log x + x^{2} (2^{x}) log x + x^{2} 2^{x} (log x).
= 2×2^{x} log x +x^{2} 2^{x} log a log x + x^{2} 2^{x} (1/x)
= x 2^{x}[log x^{2} + x log x log 2 + 1]
Sol:
f’ (x) = . log 7 ^{ }(x^{3} + 3x)
(iv) f(x) = log (sec x + tan x)
Sol:
Given, f(x) = log (sec x + tan x)
= sec x
Find the derivative of the following functions
(i) f(x) = e^{x} (x^{2} + 1)
Sol:
Given f(x) = e^{x} (x^{2} + 1)
f’ (x) = e^{x} (x^{2} + 1) + (x^{2} + 1) ^{ }(e^{x})
= e^{x} (2x + 0) + (x^{2} + 1) e^{x}
= e^{x} (x^{2} + 2x + 1)
= e^{x} (x + 1)^{2}
(iii) cos (log x + e^{x})
(iv) x = tan (e^{-y})
e^{-y} = tan^{-1} x
(v) cos [log (cot x)]
(vi) sin[tan^{-1}(e^{x})]
(vii) cos^{-1}(4x^{3} – 3x)
let y = cos^{-1}(4x^{3} – 3x)
put x = cos θ ⟹ θ = cos^{-1} x
y = cos^{-1}(4 cos^{ 3} θ – 3cos θ)
= cos^{-1}(cos 3θ)
= 3 θ
= 3 cos^{-1} x
Find f’ (x), If f(x) = (x^{3} + 6 x^{2} + 12x – 13)^{100}.
Sol:
Given f(x) = (x^{3} + 6 x^{2} + 12x – 13)^{100}
f’ (x) = 100(x^{3} + 6 x^{2} + 12x – 13)^{99} (x^{3} + 6 x^{2} + 12x – 13)
= 100(x^{3} + 6 x^{2} + 12x – 13)^{99} (3x^{2} + 12 x + 12 – 0)
=100(x^{3} + 6 x^{2} + 12x – 13)^{99} 3 (x^{2} + 4 x + 4)
= 300 (x + 2)^{2} (x^{3} + 6 x^{2} + 12x – 13)^{99}
If f(x) = 1 + x + x^{2} + x^{3} + …. + x^{100}, then find f’ (1).
Sol:
Given f(x) = 1 + x + x^{2} + x^{3} + …. + x^{100}
f’(x) = 0 + 1 + 2x + 3 x^{2} + … 100 x^{99}
f’(1) = 1 + 2 + 3 + … + 100
= 50 × 101
= 5050
From the following functions. Find their derivatives.
Sol:
Sol:
Given y = log (cosh 2x)
If x = a cos^{3} t, y = a sin^{3} t, find
Sol:
Given If x = a cos^{3} t, y = a sin^{3} t
Differentiate f(x) with respect to g(x) for the following.
derivative of f(x) with respect to g(x) =
put x = tan θ ⟹ θ = tan^{-1} x
Sol:
The post Differentiation(2m Q & S) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>
Limits
Question 1
Sol:
= 9
Question 2
Sol:
= a + a = 2a
Question 3
Sol:
Question 4
Sol:
we know that if x > 0
As x → 0+ ⟹ x > 0
As x → 0+ ⟹ x < 0
Question 5
If f (x) = , then find and . Does exist?
Sol:
Question 6
Sol:
As x → 2– ⟹ x < 2
x – 2 < 0
Question 7
Sol:
As x → 2+ ⟹ x > 2
As x → 2– ⟹ x < 2
Question 8
Sol:
Question 9
Sol:
Question 10
Sol:
Question 11
Sol:
Question 12
Sol:
Question 13
Sol:
let y = x – 1 ⟹ x = y + 1
then as x → 1, y → 0
Question 14
Sol:
Question 15
Sol:
Question 16
Sol:
As x → ∞, →0
= ∞ (3/4)
= ∞
Question 17
Sol:
We know that – 1 ≤ sin x ≤ 1
x^{2} – 1 ≤ x^{2} – sin x ≤ x^{2} + 1
Question 18
Sol:
Question 19
Sol:
Question 20
Sol:
Question 21
Sol:
Question 22
Sol:
Question 23
Sol:
The post Limits (Q’s & Ans) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>
Hyperbolic Functions
Prove that for any x∈ R, sinh (3x) = 3 sinh x + 4 sinh^{3} x
Sol:
sinh (3x) = sinh (2x + x)
= sinh 2x cosh x + cosh 2x sinh x
= (2 sinh x cosh x) cosh x + (1 + 2 sinh^{2} x) sinh x
= 2sinh x cosh^{2} x + sinh x + 2 sinh^{3} x
= 2 sinh x (1 + sinh^{2} x) + sinh x + 2 sinh^{3} x
= 2 sinh x + 2 sinh^{3} x+ sinh x + 2 sinh^{3} x
= 3 sinh x + 4 sinh^{3} x
If cosh x = , find the values of (i) cosh 2x and (ii) sinh 2x
Sol:
Cosh 2x = 2 cosh^{2} x – 1
Sinh^{2} 2x = cosh^{2} 2x – 1
If cosh x = sec θ then prove that tanh^{2}= tan^{2}
Sol:
If sinh x = 5, then show that x =
Sol:
Given, sinh x = 5
⟹ x = sinh^{-1}5
Sol:
Given tanh^{-1}
For x, y ∈ R prove that sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y)
Sol:
R.H.S = sinh (x) cosh (y) + cosh (x) sinh (y)
= sinh (x + y)
For any x∈ R, prove that cosh^{4} x – sinh^{4} x = cosh 2x
Sol:
cosh^{4} x – sinh^{4} x = (cosh^{2} x)^{2} – (sinh^{2} x)^{2}
= (cosh^{2} x + sinh^{2} x) (cosh^{2} x – sinh^{2} x)
= 1. cosh 2x
= cosh 2x
Sol:
= cosh x + sinh x
If sin hx = ¾ find cosh 2x and sinh 2x.
Sol:
Given sin hx = ¾
We know that cosh^{2} x = 1 + sinh^{2} x
= 1 + (3/4)^{2}
= 1 + 9/16
= 25/16
cos hx = 5/4
cosh 2x = 2cosh^{2} x – 1
= 2(25/16) – 1
= 25/8 – 1
= 17/8
Sinh 2x = 2 sinh x cosh x
= 2 (3/4) (5/4)
= 15/8
Prove that (cosh x – sinh x)^{ n} = cosh nx – sinh nx
Sol:
∴ (cosh x – sinh x)^{ n} = cosh nx – sinh nx
The post Hyperbolic Functions (Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>
Trigonometric Ratios Up to Transformations
Find the value of sin^{2}(π/10) + sin^{2}(4π/10) + sin^{2}(6π/10) + sin^{2}(9π/10)
Sol:
sin^{2}(π/10) + sin^{2}(4π/10) + sin^{2}(6π/10) + sin^{2}(9π/10)
= sin^{2}(π/10) + sin^{2}(π/2 – π/10) + sin^{2}(π/2+ π/10) + sin^{2}(π – π/10)
= sin^{2}(π/10) + cos^{2}(π/10) + cos^{2}(π/10) + sin^{2}(π/10)
= 1 + 1 = 2
If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ
Sol:
Given sin θ = 4/5 and θ not in the first quadrant
⇒ θ in the second quadrant
⇒ cos θ < 0
cos^{2}θ = 1 – sin^{2} θ
=1 – (4/5)^{2}
= 1 – 16/25
∴cos θ = – 3/5 (∵cos θ < 0)
If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ
Sol:
Given, 3sin θ + 4 cos θ = 5
let 4 sin θ – 3cos θ = x
(3sin θ + 4 cos θ )^{2} + (4 sin θ – 3cos θ)^{2} = 5^{2} + x^{2}
9 sin^{2} θ + 16 cos^{2} θ + 12 sin θ cos θ + 16 sin^{2} θ + 9 cos^{2} θ – 12sin θ cis θ = 25 + x^{2}
25 sin^{2} θ + 25 cos^{2} θ = 25 + x^{2}
25 = 25 + x^{2}
⇒ x^{2} = 0
x = 0
∴ 4 sin θ – 3cos θ = 0
If sec θ + tan θ =, find the value of sin θ and determine the quadrant in which θ lies
Sol:
Given, sec θ + tan θ = ———— (1)
We know that sec^{2} θ – tan^{2} θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1
(1) + (2)
⇒ (sec θ + tan θ) + (sec θ – tan θ) =
(1) – (2)
⇒ (sec θ + tan θ) – (sec θ – tan θ) =
Since sec θ positive and tan θ is negative θ lies in the 4^{th} quadrant.
Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1
Sol:
cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16)
= cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16)
= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16)
= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16)
= [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4)
= 1.1.1.1
=1
If cos θ + sin θ = cos θ, then prove that cos θ – sin θ = sin θ
Sol:
( + 1) sin θ = ( + 1) ( – 1) cos θ
Find the value of 2(sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ)
Sol:
2(sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ)
= 2[(sin^{2} θ)^{3} + (cos^{2} θ)^{3}] – 3[(sin^{2} θ)^{2} + (cos^{2})^{2}
= 2[(sin^{2} θ + cos^{2} θ)^{3} – 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ)] – 3[(sin^{2} θ + cos^{2} θ)^{2} – 2 sin^{2} θ cos^{2} θ]
= 2[1 – 3 sin^{2} θ cos^{2} θ] – 3 [1 – 2 sin^{2} θ cos^{2} θ]
= 2 – 6 sin^{2} θ cos^{2} θ – 3 + 6 sin^{2} θ cos^{2} θ
= – 1
If tan 20^{0} = λ, then show that
Sol:
Given tan 20^{0} = λ
If sin α + cosec α = 2, find the value of sin^{n} α + cosec^{n} α, n∈ Z
Sol:
Given sin α + cosec α = 2
⇒ sin α + 1/ sin α = 2
sin^{2} α + 1= 2 sin α
sin^{2} α – 2 sin α + 1= 0
(sin α – 1 )^{2} = 0
⇒ sin α – 1 = 0
sin α = 1 ⇒ cosec α = 1
sin^{n} α + cosec^{n} α = (1)^{n} + (1)^{n} =1 + 1 =2
∴ sin^{n} α + cosec^{n} α = 2
Evaluate sin^{2} + cos^{2} – tan^{2}
Sol:
Find the value of sin 330^{0}. cos 120^{0} + cos 210^{0}. Sin 300^{0}
Sol:
sin 330^{0}. cos 120^{0} + cos 210^{0}. Sin 300^{0}
=sin (360^{0} – 30^{0}). cos (180^{0} – 60^{0}) + cos (180^{0} + 30^{0}). sin (360^{0} – 60^{0})
= (– sin 30^{0}). (– cos 60^{0}) + (– cos30^{0}). (– sin60^{0})
= sin 30^{0}. cos 60^{0} + cos30^{0}. Sin60^{0}
= sin (60^{0} + 30^{0}) = sin 90^{0}
=1
Prove that cos^{4} α + 2 cos^{2} α = (1 – sin^{4} α)
Sol:
= cos^{4} α + 2 cos^{2} α (1 – cos^{2} α)
= (cos^{2} α)^{2} + 2 (1 – sin^{2} α) (sin^{2} α)
= (1 – sin^{2} α)^{2} + 2 sin^{2} α – 2sin^{4} α
= 1 + sin^{4} α – 2 sin^{2} α + 2 sin^{2} α – 2sin^{4} α
= 1 – sin^{4} α
Eliminate θ from x = a cos^{3} θ and y = b sin^{3} θ
Sol:
Given x = a cos^{3} θ and y = b sin^{3} θ
cos^{3} θ = x/a and sin^{3} θ = y/b
cos θ = (x/a)^{1/3} and sin θ = (y/b)^{1/3}
we know that sin^{2} θ + cos^{2} θ = 1
⇒ [(y/b)^{1/3}]^{2} + [(x/a)^{1/3}]^{2} = 1
(x/a)^{2/3} + (y/b)^{2/3} = 1
Find the period of the following functions
Sol:
(i) f(x) = tan 5x
we know that period of tan kx =
period of given function is = LCM (8, 6) = 24
(iv) f(x) = tan (x + 4x + 9x +…. + n^{2}x)
f(x) = tan (x + 4x + 9x +…. + n^{2}x)
= tan (1 + 4 + 9 + … + n^{2}) x
we know that period of tan kx =
Prove that sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0} =
Sol:
We know that sin^{2} A – sin^{2}B = sin (A +B) sin (A – B)
⇒ sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0}
= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)
= sin 75^{0} sin 30^{0}
∴ sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0} =
Prove that tan 70^{0} – tan20^{0} = 2 tan 50^{0}
Sol:
50^{0} = 70^{0} – 20^{0}
Tan 50^{0 }= tan (70^{0} – 20^{0})
⇒ tan 70^{0} – tan 20^{0} = tan 50^{0} (1 + tan70^{0} tan 20^{0})
tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + tan70^{0} cot (90^{0} – 20^{0})]
tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + tan70^{0} cot 70^{0}]
tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + 1]
∴ tan 70^{0} – tan20^{0} = 2 tan 50^{0}
If sin α = , sin β = and α, β are acute, show that α + β =
Sol:
tan α = 1/3 tan β = ½
tan (α + β) = 1
Sol:
Sol:
(on dividing numerator and denominator by cos 9^{0})
= tan (45^{0} + 9^{0})
= tan 54^{0}
= tan (90^{0} – 36^{0})
= cot 36^{0}
Show that cos 42^{0} + cos 78^{0} + cos 162^{0} = 0
Sol:
cos 42^{0} + cos 78^{0} + cos 162^{0}
= cos (60^{0} – 18^{0}) + cos (60^{0} + 18^{0}) + cos (180^{0} – 18^{0})
=cos 60^{0} cos18^{0} + sin 60^{0} sin 18^{0} + cos 60^{0} cos 18^{0} – sin 60^{0} sin 18^{0} – cos 18^{0}
= 2 cos 60^{0} cos 18^{0} – cos 18^{0}
= 2 (1/2) cos 18^{0} – cos 18^{0}
= cos 18^{0} – cos 18^{0}
= 0
Express sin θ + cos θ as a single of an angle
Sol:
sin θ + cos θ = 2( sin θ + cos θ)
= 2(cos 30^{0} sin θ + sin 30^{0} cos θ)
= 2 sin (θ + 30^{0})
Find the maximum and minimum value of the following functions
(i) 3 sin x –4 cos x
a= 3, b = –4 and c = 0
= 5
∴ minimum value = –5 and maximum value = 5
(ii) cos (x + ) + 2 sin (x + ) – 3
a= 1, b = 2 and c = – 3
∴ minimum value = –6 and maximum value = 0
Question 23
Sol:
Given f(x) = 7 cos x – 24sin x + 5
a= 7, b = –24 and c = 5
∴ Range = [–20, 30]
Prove that sin^{2}α + cos^{2} (α + β) + 2 sin α sin β cos (α + β) is independent of α
Sol:
sin^{2}α + cos^{2} (α + β) + 2 sin α sin β cos (α + β)
= sin^{2}α + cos (α + β) [ cos (α + β) +2 sin α sin β]
= sin^{2}α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]
=sin^{2}α + cos (α + β) [ cos α cos β + sin α sin β]
=sin^{2}α + cos (α + β) cos (α –β)
= sin^{2} α + cos^{2} α – sin^{2} β
=1 – sin^{2} β
= cos^{2} β
Sol:
= tan θ
For what values of x in the first quadrant is positive?
Sol:
⟹ 0 < 2x < π/2 (∵ x is in first quadrant)
⟹ 0 < x < π/4
If cos θ = and π < θ < 3π/2, find the value of tan θ/2.
Sol:
π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4
tan θ/2 < 0
= – 2
If A is not an integral multiple of π/2, prove that cot A – tan A = 2 cot 2A.
Sol:
= 2 cot 2A
Evaluate 6 sin 20^{0} – 8sin^{3} 20^{0}
Sol:
6 sin 20^{0} – 8sin^{3} 20^{0} = 2 (3 sin 20^{0} – 4sin^{3} 20^{0})
= 2 sin 3(20^{0})
= 2 sin 60^{0}
Express cos^{6} A + sin^{6} A in terms of sin 2A.
Sol:
cos^{6} A + sin^{6} A
= (sin^{2} A)^{3} + (cos^{2} A)^{3}
= (sin^{2} A + cos^{2} A)^{3} – 3 sin^{2} A cos^{2} A (sin^{2} A + cos^{2} A)
= 1 – 3 sin^{2} A cos^{2} A
=1 – ¾ (4 sin^{2} A cos^{2} A)
= 1 – ¾ sin^{2}2 A
If 0 < θ < π/8, show that = 2 cos (θ/2)
Sol:
=2 cos (θ/2)
Find the extreme values of cos 2x + cos^{2}x
Sol:
cos 2x + cos^{2}x = 2cos^{2} x– 1 + cos^{2} x
=3cos^{2} x – 1
We know that – 1 ≤ cos x ≤ 1
⟹ 0 ≤ cos^{2} x ≤ 1
3×0 ≤ 3×cos^{2} x ≤ 3×1
0– 1 ≤3 cos^{2} x – 1≤ 3– 1
– 1≤3 cos^{2} x – 1≤ 2
Minimum value = – 1
Maximum value = 2
Sol:
= 4
Prove that sin 78^{0} + cos 132^{0} =
Sol:
sin 78^{0} + cos 132^{0} = sin 78^{0} + cos (90^{0} + 42^{0})
= sin 78^{0} – sin 42^{0}
= 2 cos 60^{0} sin 18^{0}
Find the value of sin 34^{0} + cos 64^{0} – cos4^{0}
Sol:
sin 34^{0} + cos 64^{0} – cos4^{0}
= sin 34^{0} – 2sin 34^{0} sin 30^{0}
= sin 34^{0} – 2 sin 34^{0} (1/2)
=sin 34^{0} – sin 34^{0}
=0
Prove that 4(cos 66^{0} + sin 84^{0}) =
Sol:
4(cos 66^{0} + sin 84^{0})
=4(cos 66^{0} + sin (90^{0 }– 6^{0})
=4(cos 66^{0} + cos (6^{0})
=8 cos 36^{0} cos 30^{0}
Prove that (tan θ + cot θ)^{2} = sec^{2} θ + cosec^{2} θ = sec^{2} θ. cosec^{2} θ
Sol:
= sec θ. cosec θ
(tan θ + cot θ)^{2} = sec^{2} θ. cosec^{2} θ
= sec^{2} θ. cosec^{2} θ
The post Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>
Question 1
If a = 6i +2 j +3 k, b = 2i – 9 j+ 6k, then find the angle between the vectors a and b
Sol:
Given vectors are a = 6i +2 j +3 k, b = 2i – 9 j+ 6k
If θ is the angle between the vectors a and b, then cos θ =
a .b = (6i +2 j +3 k). (2i – 9 j+ 6k) = 6(2) + 2 (– 9) + 3(6)
= 12 – 18 + 18 = 12
= 7
= 11
Question 2
If a = i +2 j –3 k, b = 3i – j+ 2k, then show that a + b and a – b are perpendicular to each other.
Sol:
Given vectors are a = i +2 j –3 k, b = 3i – j+ 2k
a + b = (i +2 j –3 k) + (3i – j+ 2k) = 4i + j – k
a – b = (i +2 j –3 k) – (3i – j+ 2k) = –2i +3 j – 5k
(a + b). (a – b) = (4i + j – k). (–2i +3 j – 5k)
= – 8 + 3 + 5
= 0
∴ a + b and a – b is perpendicular to each other.
If a and b be non-zero, non-collinear vectors. If , then find the angle between a and b
Sol:
Squaring on both sides
(a + b) (a + b) = (a – b) (a – b)
a^{2} + 2 a. b + b^{2} = a^{2} – 2 a.b + b^{2}
⟹ 4 a.b = 0
a.b = 0
∴ the angle between a and b is 90^{0}
If = 11, = 23 and = 30, then find the angle between the vectors a and b and also find
Sol:
(11)^{2} – 2 ×11×23 cos θ + (23)^{2} = 900
121 – 506 cos θ + 529 = 900
650 – 506 cos θ = 900
= (11)^{2} + 2 ×11×23 cos θ + (23)^{2}
= 121 + 2 ×11×23 × + 529
= 400
If a = i – j – k and b = 2i – 3j + k, then find the projection vector of b on a and its magnitude.
Sol:
Given vectors are a = i – j – k and b = 2i – 3j + k
a.b = (i – j – k). (2i – 3j + k) = 2 + 3 – 1 = 4
The projection vector of b on a =
The magnitude of the projection vector = =
If the vectors λ i – 3j + 5k and 2λ i – λ j – k are perpendicular to each other, then find λ
Sol:
let a = λ i – 3j + 5k, b = 2λ i – λ j – k
Given, that a and b are perpendicular to each other
⟹ a.b = 0
(λ i – 3j + 5k). (2λ i – λ j – k) = 0
2 λ^{2} + 3 λ – 5 = 0
2 λ^{2} + 5 λ – 2 λ – 5 = 0
λ (2 λ + 5) – 1 (2 λ + 5) = 0
(2 λ + 5) ((λ – 1) = 0
λ = 1 or λ = -5/2
Find the Cartesian equation of the plane passing through the point (– 2, 1, 3) and perpendicular to the vector 3i + j + 5k
Sol:
let P (x, y, z) be any point on the plane
⟹ OP = xi + yj + zk
OA = – 2i +j +3k
AP = OP – OA = (xi + yj + zk) – (– 2i +j +3k)
AP = (x + 2) i + (y – 1) j + (z – 3) k
AP is perpendicular to the vector 3i + j + 5k
⟹ 3 (x + 2) + (y – 1) + 5(z – 3) = 0
⟹ 3x + 6 + y – 1 + 5z – 15 = 0
∴ 3x + y + 5z – 10 = 0 is the required Cartesian equation of the plane
Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4
Sol:
Given plane equations are: 2x – 3y – 6z = 5,6x + 2y – 9z = 4
Vector equations of the above planes are: r. (2i – 3j – 6k) = 5 and r. (6i + 2j – 9k) = 4
⟹ n_{1} = 2i – 3j – 6k and n_{2} = 6i + 2j – 9k
If θ is the angle between the planes r. n_{1} = d_{1} and r. n_{2} = d_{2}, then
a = 2i – j + k, b = i – 3j – 5k. Find the vector c such that a, b, and c form the sides of a triangle.
Sol:
Given a = 2i – j + k, b = i – 3j – 5k
If a, b, and c form the sides of a triangle, then a + b + c = 0
⟹ a + b = – c
⟹ c = – (a + b)
= – [(2i – j + k) +( i – 3j – 5k)]
= – (3i –4 j –4k)
∴ c = – 3i +4 j + 4k
Find the equation of the plane through the point (3, –2, 1) and perpendicular to the vector (4, 7, –4).
Sol:
Let a = 3i – 2j + k and n = 4i + 7j – 4k
The equation of the plane passing through point A(a) and perpendicular to the vector n is (r – a). n = 0
⟹ [r – (3i – 2j + k)]. (4i + 7j – 4k) = 0
⟹ r. (4i + 7j – 4k)– [(3i – 2j + k). (4i + 7j – 4k)] = 0
r. (4i + 7j – 4k)– (12 – 14 – 4) = 0
r. (4i + 7j – 4k)– 6 = 0
r. (4i + 7j – 4k) = 6
Find the unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k
Sol:
The vector which is parallel to the XOY-plane is of the form xi + yj
The vector which is parallel to the XOY-plane and perpendicular to 4i – 3j + k
is 3i + 4j
∴ The unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k =
If a + b + c = 0, = 3, = 5 and = 7, then find the angle between a and b
Sol:
Given, a + b + c = 0, = 3, = 5 and = 7
a + b = – c
3^{2} + 5^{2} + 2 cos θ = 7^{2}
9 + 25 + 2.3.5 cos θ = 49
34 + 30cos θ = 49
30cos θ = 49 – 34
30cos θ = 15
cos θ = 15/30 = 1/2
∴ θ = π/3
If a = 2i – 3j + 5k, b = – i + 4j + 2k, then find a × b and unit vector perpendicular to both a and b
Sol:
Given, a = 2i – 3j + 5k, b = – i + 4j + 2k
= i (–6 – 20) – j (4 + 5) + k (8 – 3)
= –26i – 9j + 5k
The unit vector perpendicular to both a and b =
If a = i + j + 2k and b = 3i + 5j – k are two sides of a triangle, then find its area.
Sol:
Given, a = i + 2j + 3k and b = 3i + 5j – k
If a, b are two sides of a triangle, then area of the triangle =
= i (–2 – 15) – j (–1 – 9) + k (5 – 6)
= –17i + 10 j – k
Find the area of the parallelogram for which the vectors a = 2i – 3j and b = 3i – k are adjacent sides.
Sol:
Given, a = 2i – 3j and b = 3i – k are adjacent sides of a parallelogram
The area of the parallelogram whose vectors a , b are adjacent sides =
= i (3 – 0) – j (–2 – 0) + k (0 + 9)
=3 i +2 j +9 k
∴ The area of the parallelogram =
Let a, b be two non-collinear unit vectors. If α = a – (a . b) b and β = a × b, then show that
Sol:
= 1 – cos^{2} θ
= sin^{2} θ
= 1 + cos^{2} θ – 2cos^{2} θ
= 1– cos^{2} θ
= sin^{2} θ
Sol:
Let a = xi + yj + zk
= i ( 0 – 0) – j (0 – z) + k (0 – y)
= – yk + zj
Similarly, = x^{2} + z^{2} and = y^{2} + x^{2}
+ + = y^{2} + z^{2} + x^{2} + z^{2} + y^{2} + x^{2}
= 2(x^{2} + y^{2} +z^{2})
If = 2, = 3 and (p, q) = , then find
Sol:
= 2 × 3 sin
= 2 × 3×1/2
= 3
If 4i + j + pk is parallel to the vector i + 2j + 3k, then find p.
Sol:
Given 4i + j + pk is parallel to the vector i + 2j + 3k
⇒ p = 12
Compute a× (b + c) + b× (c + a) + c× (a + b)
Sol:
a× (b + c) + b× (c + a) + c× (a + b)
= a× b + a× c + b× c + b× a + c × a + c × b
= a× b + a× c + b× c –a × b – a × c – b ×c
= 0
Compute 2j× (3i – 4k) + (i + 2j) × k
Sol:
2j× (3i – 4k) + (i + 2j) × k
= 6(j × i) – 8(j × k) + (i × k) + 2(j × k)
= – 6k – 8i – j + 2i
= – 6i –j –6 j
The post Product of Vectors (Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>
Addition of Vectors
Find the unit vector in the direction of
The unit vector in the direction of a vector is given by
Find a vector in the direction of a where that has a magnitude of 7 units.
The unit vector in the direction of a vector is
The vector having the magnitude 7 and in the direction of is
Find the unit vector in the direction of the sum of the vectors, a = 2i + 2j – 5k and b = 2i + j + 3k
Sol Given vectors are a = 2i + 2j – 5k and b = 2i + j + 3k
a + b = (2i + 2j – 5k) + (2i + j + 3k) = 4i + 3j – 2k
Write the direction cosines of the vector
QUESTION 5
Show that the points whose position vectors are – 2a + 3b + 5c, a + 2b + 3c, 7 a – c are collinear when a, b, c are non-collinear vectors
Sol: Let OA = – 2a + 3b + 5c, OB = a + 2b + 3c, OC = 7 a – cA
B = OB – OA = a + 2b + 3c – (– 2a + 3b + 5c)
AB = 3a – b – 2c
AC = OC – OA = 7 a – c – (– 2a + 3b + 5c)
AC = 9a – 3b – 6c = 3(3a – b – 2c)
AC = 3 AB
A, B and C are collinear
ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) AL and AM in terms of AB and AD (ii) 𝛌, if AM = 𝛌 AD – LM
Sol: Given, ABCD is a parallelogram and L and M are middle points of BC and CD
(i) Take A as the origin
M is the midpoint of CD
= AD + ½ AB (∵ AB = DC)
L is the midpoint of BC
= AB + ½ AD ((∵ BC = AD)
(ii) AM = 𝛌 AD – LM
AM + LM= 𝛌 AD
AD + ½ AB + AD + ½ AB – (AB + ½ AD) = 𝛌 AD
AD + ½ AB + AD + ½ AB – AB – ½ AD = 𝛌 AD
3/2 AD = 𝛌 AD
∴𝛌 = 3/2
If G is the centroid of the triangle ABC, then show that OG = when, are the position vectors of the vertices of triangle ABC.
Sol: OA = a, OB = b, OC = c and OD = d
D is the midpoint of BC
G divides median AD in the ratio 2: 1
If = , = are collinear vectors, then find m and n.
Sol: Given , are collinear vectors
Equating like vectors
2 = 4 λ; 5 = m λ; 1 = n λ
∴ m = 10, n = 2
Let If , . Find the unit vector in the direction of a + b.
The unit vector in the direction of a + b =
If the vectors – 3i + 4j + λk and μi + 8j + 6k. are collinear vectors, then find λ and μ.
Sol: let a = – 3i + 4j + λk, b = μi + 8j + 6k
⟹ a = tb
– 3i + 4j + λk = t (μi + 8j + 6k)
– 3i + 4j + λk = μt i + 8t j + 6t k
Equating like vectors
– 3 = μt; 4 = 8t, λ = 6t
4 = 8t
∴ μ=– 6, λ = 3
ABCD is a pentagon. If the sum of the vectors AB, AE, BC, DC, ED and AC is 𝛌 AC then find the value of 𝛌
Sol: Given, ABCD is a pentagon
AB + AE + BC + + DC + ED + AC = 𝛌 AC
(AB + BC) + (AE + DC + ED) + AC = 𝛌 AC
AC + AC + AC = 𝛌 AC
3 AC = 𝛌 AC
𝛌 = 3
If the position vectors of the points A, B and C are – 2i + j – k and –4i + 2j + 2k and 6i – 3j – 13k respectively and AB = 𝛌 AC, then find the value of 𝛌
Sol: Given, OA = – 2i + j – k , OB = –4i + 2j + 2k and OC = 6i – 3j – 13k
AB = OB – OA = –4i + 2j + 2k – (– 2i + j – k)
= –4i + 2j + 2k +2i – j + k
= –2i + j + 3k
AC = OC – OA = 6i – 3j – 13k – (– 2i + j – k)
= 6i – 3j – 13k +2i – j + k
= 8i –4 j –12k
= – 4 (2i + j + 3k)
AC = – 4 AB
Given AB = 𝛌 AC
𝛌 = – 1/4
If OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k then find the vector OD
Sol: Given OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k
OD = OA + AB + BC + CD
= i + j +k + 3i – 2j + k + i + 2j – 2k + 2i + j +3k
= 7i + 2j +4k
Let a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k, then find the unit vector in the opposite direction of a + b + c
Sol: Given, a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k
a + b + c = 2i +4 j –5 k + i + j+ k + j +2 k
= 3i +6j –2k
The unit vector in the opposite direction of a + b + c is
Is the triangle formed by the vectors 3i +5j +2k, 2i –3j –5k, –5i – 2j +3k
Sol: Let a =3i +5j +2k, b = 2i –3j –5k, c = –5i – 2j +3k
∴ Given vectors form an equilateral triangle.
Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are a, b, (3a – 2b) are collinear.
Sol: the vector equation of the straight line passing through two points a, b is
r = (1 – t) a+ t b
3a – 2b = (1 – t) a+ t b
Equating like vectors
1 – t = 3 and t = – 2
∴ Given points are collinear.
OABC is a parallelogram If OA = a and OC = c, then find the vector equation of the side BC
Sol: Given, OABC is a parallelogram and OA = a, OC = c
The vector equation of BC is a line which is passing through C(c) and parallel to OA
⟹ the vector equation of BC is r = c + t a
If a, b, c are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A
Sol: Given OA = a, OB = b, OC = c
D is mid of BC
The equation of AD is
Find the vector equation of the line passing through the point 2i +3j +k and parallel to the vector 4i – 2j + 3k
Sol: Let a =2i +3j +k, b = 4i – 2j + 3k
The vector equation of the line passing through a and parallel to the vector b is r = a + tb
r = 2i +3j +k + t (4i – 2j + 3k)
= (2 + 4t) i + (3 – 2t) j + (1 + 3t) k
Find the vector equation of the plane passing through the points i – 2j + 5k, – 2j –k and – 3i + 5j
Sol: The vector equation of the line passing through a, b and cis r = (1 – t – s) a + tb + sc
⟹ r = (1 – t – s) (i – 2j + 5k) + t (– 2j –k) + s (– 3i + 5j)
= (1 – t – 4s) i + (– 2 – 3t + 7s) j + (5 – 6t – 5s) k
The post Addition of Vectors (Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>
Matrices
If A = , then show that A^{2} = –I
∴ A^{2} = –I
If A = , and A^{2} = 0, then find the value of k.
A^{2} = 0
8 + 4k = 0, – 2 – k = 0 and –4 + k^{2} = 0
4k = –8; k = –2; k^{2} = 4
k = –2; k = –2; k = ± 2
∴ k =– 2
Trace of A = 1 – 1 + 1 = 1
If A = , B = and 2X + A = B, then find X.
Sol: Given A = , B = and 2X + A = B
2X = B – A
Find the additive inverse of A, If A =
Additive inverse of A = – A
If , then find the values of x, y, z and a.
⟹ x- 1 = 1 – x ; y – 5 = – y ; z = 2 ; 1 + a = 1
⟹ x + x = 1 + 1; y + y = 5; z = 2; a =1– 1
⟹ 2x = 1; 2y = 5; z = 2; a = 0
∴ x = ½ ; y = 5/2; z = 2; a = 0
Construct 3 × 2 matrix whose elements are defined by a_{ij} =
Sol:
Let A= _{ }
a_{11} = 1
a_{22} = 2
a_{31} = 0
If A = and B = , do AB and BA exist? If they exist, find them. BA and AB commutative with respect to multiplication.
Sol: Given Matrices are A = B =
Order of A = 2 × 3 and Order of B = 3 × 2
AB and BA exist
AB and BA are not Commutative under Multiplication
Define Symmetric and Skew Symmetric Matrices
Sol:
Symmetric Matrix: Let A be any square matrix, if A^{T} = A, then A is called Symmetric Matrix
Skew Symmetric Matrix: Let A be any square matrix if A^{T} = –A, then A is called Skew Symmetric Matrix
If A = is a symmetric matrix, then find x.
Sol: Given, A = is a symmetric matrix
⟹ A^{T} = A
⟹ x = 6
If A = is a skew-symmetric matrix, then find x
Sol: Given A = is a skew-symmetric matrix
⟹ A^{T} = – A
⟹ x = –x
x+ x = 0 ⟹ 2x = 0
⟹ x = 0
If A = and B = , then find (A B^{T})^{ T}
If A = and B = , then find A + B^{T}
If A = , then show that AA^{T} = A^{T}A = I
∴ AA^{T} = A^{T}A = I
Find the minor of – 1 and 3 in the matrix
Sol: Given Matrix is
Find the cofactors 0f 2, – 5 in the matrix
Sol: Given matrix is
Cofactor of 2 = (–1)^{2 + 2} = –3 + 20 = 17
Cofactor of – 5 = (–1)^{3 + 2} = –1(2 – 5) = –1(–3) = 3
If ω is a complex cube root of unity, then show that = 0(where 1 + ω+ω^{2} = 0)
R_{1} → R_{1} + R_{2} + R_{3}
If A = and det A = 45, then find x.
Det A = 45
⟹ 1(3x + 24) – 0 (2x – 20) + 0 (– 12 – 15) = 45
⟹ 3x + 24 = 45
3x = 45 – 24
3x = 21
x = 7
Find the adjoint and inverse of the following matrices
(i)
(ii)
Det A = a (bc – 0) – 0(0 – 0) + 0(0 – 0)
Det A = abc ≠ 0
Adj A = (Cofactor matrix of A)^{ T}
Find the rank of the following matrices.
Det A = 1 (0 – 2) – 2(1 – 0) + 1(– 1 – 0)
= – 2– 2– 1
= – 5 ≠ 0
∴ Rank of A = 3
Det A = – 1 (24 – 25) + 2(18 – 20) + – 3(15 – 16)
= – 1– 4 + 3
= – 0
Det B = – 4 + 6 = 2 ≠ 0
∴ Rank of A = 2
Det of Sub matrix of A = – 1 – 0 = – 1 ≠ 0
∴ Rank of A = 2
Det of Sub matrix of A =1 (1 – 0) – 0(0 – 0) + 0(0 – 0)
= 1≠ 0
∴ Rank of A = 3
The post Matrices ( Qns & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>
The Plane (2m Questions & Solutions) || V.S.A.Q’S|| Read More »
The post The Plane (2m Questions & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (2, 3, – 5).
Sol:
The plane passes through A and perpendicular to OA, then the line segment OA is normal to the plane.
Dr’s of OA = (2, 3, – 5)
The equation of the plane passing through point (x_{1}, y_{1}, z_{1}) and dr’s (a, b, c) is
a(x – x_{1}) + b (y – y_{1}) + c (z – z_{1}) = 0
⟹ 2(x – 2) + 3 (y – 3) – 5 (z + 5) = 0
2x – 4 + 3y – 9 – 5z – 25 = 0
2x + 3y – 5z – 38 = 0
Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4)
Sol:
The equation of the plane passing through the points (x_{1}, y_{1}, z_{1}) (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}) is
The plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4) is
x (30 – 20) – (y + 1) (20 – 6) + (z + 1) (40 – 18) = 0
x (10) – (y + 1) (14) + (z + 1) (22) = 0
10x – 14y – 14 + 22z + 22 = 0
10x – 14y + 22z + 8 = 0
2(5x – 7y + 11z + 4) = 0
∴ the equation of the plane is 5x – 7y + 11z + 4 = 0
Find the equation to the plane parallel to the ZX-plane and passing through (0, 4, 4).
Sol:
Equation of ZX-plane is y = 0
The equation of the plane parallel to the ZX-plane is y = k
But it is passing through (0, 4, 4)
⟹ y = 4
Find the equation to the plane passing through the point (α, β, γ) and parallel to the plane axe + by + cz + d = 0.
Sol:
The equation of the plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0
But it is passing through the point (α, β, γ)
a α + b β + c γ + k = 0
⟹ k = – a α – b β – c γ
The equation of the plane is ax + by + cz – a α – b β – c γ = 0
⟹ a(x – α) + b (y – β)+ c (z – γ) = 0
Find the angle between the planes 2x – y + z = 6 and x + y + 2z = 7.
Sol: If θ is the angle between the planes a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 and a_{2}x + b_{2} y + c_{2} z + d_{2} = 0, then cos θ =
Cos θ = cos 60^{0}
θ = 60^{0} =
Reduce the equation x + 2y – 2z – 9 = 0 to the normal form and hence find the dc’s of the normal to the plane.
Sol: Given plane is x + 2y – 2z – 9 = 0
x + 2y – 2z = 9
dc’s of the normal to the plane are
Suppose a plane makes intercepts 2, 3, 4 on X, Y, Z axes respectively. Find the equation of the plane in the intercept form.
Sol: Given a = 2, b = 3, c = 4
The equation of the line in the intercept form is
Express x – 3y + 2z = 9 in the intercept form
Sol: Given plane is x – 3y + 2z = 9
a = 9, b = – 3, c = 9/2
Find the direction cosine of the normal to the plane x + 2y + 2z – 4 = 0.
Sol: Given plane is x + 2y + 2z – 4 = 0
We know that Dr’s of the normal to the plane ax + by + cz + d = 0 are (a, b, c)
⟹ dc’s of the normal to the plane =
⟹ dr’s of the normal to the plane x + 2y + 2z – 4 = 0 are (1, 2, 2)
⟹ dc’s of the normal to the plane are
Find the midpoint of the line joining the points (1, 2, 3) and (–2, 4, 2)
Sol: Given points are A (1, 2, 3), B (–2, 4, 2)
Show that the points A ( – 4, 9, 6), B (– 1, 6, 6), and C(0, 7, 10) right-angled isosceles triangle.
Sol:
Distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is PQ =
AB^{2} + BC^{2 } = 18 + 18 = 36 = AC^{2}
∴ ∠ B = 90^{0}
points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle
Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 = 0
Sol:
Let P (x, y, z) be the locus of the point
A (0, y, 0) be any point on Y – axis
B = (1, 2, – 1)
Condition is PA = 3PB
PA^{2} = (3PB)^{2}
z^{2} = 9[x^{2 }– 2x + 1 + y^{2} – 4y + 4 + z^{2} + 2z + 1]
x^{2} + z^{2} = 9x^{2 }– 18x + 9 + 9y^{2} – 36y + 36 +9 z^{2} + 18z + 9
∴ 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 =0
A, B, C are three points on OX, OY and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C and O
Sol:
Let P (x, y, z) be the required point
O = (0, 0, 0) A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)
Given, AP = BP = CP = OP
AP = OP
⟹ AP^{2} =OP^{2}
(x – a )^{2 }+ y^{2} + z^{2} = x^{2} + y^{2} + z^{2}
x^{2} + a^{2} – 2ax + y^{2} + z^{2} = x^{2} + y^{2} + z^{2}
a^{2} – 2ax = 0
a (a – 2x) = 0
a – 2x = 0 (∵ a≠0)
a = 2x ⟹ a/2
Similarly, y = b/2 and z = c/2
∴ P = (a/2, b/2, c/2)
Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear
Sol:
Given points are A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2)
AB + BC = AC
∴ A B and C are collinear
Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.
Sol:
Let A = (5, – 1, 7), B = (x, 5, 1)
Given AB = 9
⟹ AB^{2} = 81
(5 – x)^{2} + (– 1 – 5)^{2} + (7 – 1)^{2} = 81
(5 – x)^{2 }+ 36 + 36 = 81
(5 – x)^{2} + 72 = 81
(5 – x)^{2 }= 81 – 72 = 9
(5 – x) = ± 3
5 – x = 3 or 5 – x = – 3
5 – 3 = x or 5 + 3 = x
x = 2 or x = 8
If the point (1, 2, 3) is changed to the point (2, 3, 1) through the translation of axes. Find a new origin.
Sol:
Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)
x = X + h, y = Y + k, z = Z + l
h = x – X, k = y – Y, l = z – Z
h = 1 – 2, k = 2 – 3, l = 3 – 1
h = – 1, k = – 1, l = 2
New origin is (– 1, – 1, 2)
By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.
Sol:
If a point P divides the line segment joining the points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) in the ratio, then P =
Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.
Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).
Sol:
Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)
Suppose P divides AB in the ratio k : 1
7 – 2k = k + 1
7 – 1 = k + 2k
6 = 3k
k = 2
∴ P divides AB in the Ratio 1 : 2
Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not.
Sol:
Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)
Let C divides AB in the ratio k : 1
k = 4 (k + 1)
2 – 4k = 4k + 4
– 4k– 4k = 4 – 2
– 8k = 2
K = -1/4
C divides AB in the Ratio 1 : 4 externally
∴ A, B, C are collinear
Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)
Sol:
The centroid of the triangle whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) is
∴ The centroid of the triangle =
Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)
Sol:
The centroid of the triangle whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is
the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is
Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)
Sol:
Let P be any point on the YZ-plane
P = (0, y, z)
Let P divides AB in the ratio k:1
3k + 2 = 0
3k =– 2
YZ-plane divides AB in the ratio – 2:3
Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).
Sol:
let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)
ABCD is a parallelogram
The midpoint of AC = Midpoint of BD
⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0
x = 3, y = 3, z = 1
∴ The fourth vertex D = (3, 3, 1)
A (5, 4, 6), B (1, –1, 3), C (3, 3, 1) are three points. Find the coordinates of the point in which the bisector of ∠BAC meets the side BC.
Sol:
We know that the bisector of ∠BAC divides BC in the ratio AB:AC
= 5:3
If D is a point where the bisector of ∠BAC meets BC
⟹ D divides BC in the ratio 5:3
If M (α, β, γ) is the midpoint of the line joining the points (x_{1}, y_{1}, z_{1}) and B, then find B
Sol:
Let B (x, y, z) be the required point
M is the midpoint of AB
⟹ 2 α = x + x_{1}; 2 β = y + y_{1}; 2 γ = z + z_{1}
x =2 α – x_{1}; x =2 β – y_{1;} x =2 γ – z_{1}
_{ }∴ B = (2 α – x_{1}, 2 β – y_{1}, 2 γ –)
If H, G, S and I respectively denote orthocenter, centroid, circumcenter and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.
Sol:
Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)
AB = BC = AC
⟹ ∆ ABC is an equilateral triangle
We know that, in an equilateral triangle orthocenter, centroid, circumcenter and incentre are same
= (2, 2, 2)
∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)
Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).
Sol:
Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)
a = 5
b = 4
c = 3
∴ I = (1, 1,0)
Find the ratio in which the point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also, find the harmonic conjugate of P
Sol:
Harmonic Conjugate: If P divides AB in the ratio m: n, then the Harmonic Conjugate of P (i.e., Q) divides AB in the ratio –m: n.
Given points are A (3, 2, – 4), B (9, 8, –10) and P (5, 4, – 6)
P divides AB in the ratio is x_{2} – x : x – x_{2}
= 3 – 5 : 5 – 9
= 1 : 2 (internally)
Let Q be the harmonic conjugate of P
⟹ Q divides AB in the ratio –1: 2
= (–3, –4, –2)
Q (–3, –4, –2) is the harmonic conjugate of P (5, 4, – 6)
If (3, 2, – 1), (4, 1,1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of the tetrahedron, find the fourth Vertex.
Sol:
Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)
Let the fourth vertex is D = (x, y, z)
The centroid of the tetrahedron whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is
13 + x = 16, 5 + y = 8, 5 + z = 2
x = 3, y = 3, z = 3
∴ the fourth vertex D = (3, 3, 3)
Show that the points A(3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear and find the ratio in which B divides AC.
Sol:
Given points are A(3, 2, –4), (5, 4, –6) and C(9, 8, –10)
∴ the points A (3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear
B divides AB in the ratio is AB:BC = = 1:2
The post 3D Coordinates (Q’s & Ans) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>
Straight Lines (2m Questions & Solutions) || V.S.A.Q’S|| Read More »
The post Straight Lines (2m Questions & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>Prove that the points (1, 11), (2, 15), and (– 3, – 5) are collinear, and find the equation of the line containing them.
Sol:
Let A (1, 11), B (2, 15), and C (– 3, – 5)
The slope of the line segment joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is
Slope of AB = = 4
QUESTION2.
Find the condition for the points (a, 0), (h, k), and (0, b) to be collinear.
Sol:
Let A (a, 0), B (h, k) and C (0, b)
The slope of the line segment joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is
Given that A, B, and C are collinear points
The slope of AB = The slope of BC
⟹ – hk = (h – a) ( b – k)
– hk = hb – hk – ab + ak
⟹ 0 = hb + ak – ab
Find the equations of the straight lines which makes intercepts whose sum is sum is 5 and product is 6.
Sol:
The equation of the line in the intercept form is
Given that, a + b = 5 and ab = 6
⟹ b = 5 – a
a(5 – a) = 6
5a – a^{2} = 6
a^{2} – 5a + 6 = 0
a – 3a – 2a + 6 = 0
a (a – 3) – 2(a – 3) = 0
(a – 3) (a – 2) = 0
a = 3 or a = 2
case (i) if a = 3 ⟹ b = 2
case (ii) if a = 2 ⟹ b = 3
Find the equation of the straight line which makes an angle 135^{0} with the positive X – axis measured countered clockwise and passing through the point (– 2, 3).
Sol:
Slope of the line m = tan 135^{0} = – 1
The point is (– 2, 3)
The equation of the straight line in slope point form is (y – y_{1}) = m (x – x_{1})
The equation of the line passing through the point (– 2, 3) with slope – 1 is
y – 3 = – 1 (x + 2)
y – 3 =– x – 2
x + y – 1 = 0
Find the equation of the straight line passing through the points (1, – 2) and (– 2, 3).
Sol:
Given points are (1, – 2), (– 2, 3)
The equation of the straight line in two points form is (y – y_{1}) = (x – x_{1})
The equation of required straight line is
– 3 (y + 2) = 5 (x – 1)
– 3y – 6 = 5x – 5
5x + 3y + 1 = 0
QUESTION6.
Find the slopes of the line x + y = 0 and x – y = 0
Sol:
The slope of the line ax + by + c = 0 is
The slope of the line x + y = 0 is = – 1
The slope of the line x – y = 0 is = 1
Find the angle which the straight-line y = x – 4 makes with the Y-axis.
Sol:
Compare with y = mx + c
Angle made by the line with X-axis is
Angle made by the line with Y-axis is
Find the equation of the reflection of the line x = 1 in the Y-axis.
Sol:
Given equation is x = 1
Reflection about the Y-axis is x =– 1
Required equation of the line is x + 1 = 0
QUESTION9.
Write the equations of the straight lines parallel to X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis.
Sol:
(i) The equation of the straight line parallel to X-axis which is at a distance of 3 units above the X-axis is y = 3
⟹ y – 3 = 0
(ii) The equation of the straight line parallel to X-axis which is at a distance of 4 units below the X-axis is y = – 4
⟹ y + 4 = 0
QUESTION10.
Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it.
Sol:
(i) The equation of the straight line parallel to the Y-axis which is at a distance of 2 units from the Y-axis to the right of it is x = 2
⟹ x – 2 = 0
(ii) The equation of the straight line parallel to the Y-axis which is at a distance of 5 units from the Y-axis to the left of it is x =– 5
⟹ x + 5 = 0
Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2.
Sol:
Given the slope of the line passing through (2, 5) and (x, 3) is 2.
The slope of the line segment joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is
⟹ 2(x – 2) = – 2
x – 2 = – 1
x = 1
Find the equation of the straight line passing through (– 4, 5) and making non – zero intercepts on the coordinate axes whose sum is zero.
Sol:
The equation of a line in the intercept form is
Given a = b
x + y = a
but it is passing through (– 4, 5)
– 4+ 5 = a
a = 1
∴ The equation of the required straight line is x + y = 1
x + y – 1 = 0
QUESTION13.
Find the equation of the straight line passing through (– 2, 5) and cutting off equal and non – zero intercepts on the coordinate axes.
Sol:
The equation of a line in the intercept form is
Given a + b = 0
b = – a
x – y = a
but it is passing through (– 2, 4)
– 2– 4 = a
a = –6
∴ The equation of the required straight line is x – y = –6
x + y + 6 = 0
Sol:
The equation of the straight line in the normal form is x cos α + y sin α = p
p = 4 and α = 135^{0}
x cos 135^{0} + y sin 135^{0} = 4
QUESTION15.
Sol:
Given θ = 135^{0} and (h, k) = (3, 2)
The parametric equations are: x = h + r Cos θ, y = k +r Sin θ
x = 3 + r Cos 135^{0}, y = 2 + r Sin 135^{0}
(i) x + y +1 = 0
Sol:
Given equation is x + y +1 = 0
x + y =– 1
– x – y = 1
(ii) x + y = 2
Sol:
Given equation is x + y = 2
Sol:
Given equation is 3x + 4y = a
The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes is
a^{2 }= 24 ⟹ a = 12(∵ a > 0)
QUESTION18.
Find the sum of the squares of the intercepts of the line 4x – 3y = 12
Sol:
Given equation is 4x – 3y = 12
a = 3 and b =– 4
the sum of the squares of intercepts = a^{2} + b ^{2}
= 3^{2} + (– 4)^{2}
= 9 + 16 = 25
Find the angle made by the straight – line y = – x + 3 with the positive direction of the X-axis measured in the counter-clockwise direction.
Sol:
Give equation of straight line is y = x + 3
It is in the form of y = mx + c
QUESTION 20.
Find the equation of the straight line in the symmetric form, given the slope and point on the line (2, 3).
Sol:
Equation of the line in the symmetric form is
Given Point (x_{1}, y_{1}) = (2, 3) and slope m =
⟹ θ = 60^{0}
If the product of the intercepts made by the straight-line x tan α + y sec α = 1 (0≤ α < π/2) on the coordinate axes is equal to sin α, find α.
Sol:
Given equation is x tan α + y sec α = 1
a = cot α, b = cos α
product of intercepts = sin α
⟹ ab = sin α
cot α.cos α = sin α
(cos α)/ (sin α). cos α = sin α
⟹ cos^{2} α = sin^{2} α
⟹ tan^{2} α = 1
∴ α = 45^{0}
QUESTION22.
If the sum of the reciprocals of the intercepts made by a variable straight–line on the axes of coordinates is a constant, then prove that the line always passes through a fixed point.
Sol:
The equation of the line in the intercept form is
Let the sum of the reciprocals of intercepts is k
⟹
∴ the line always passes through a fixed point (1/k, 1/k).
QUESTION23.
Find the ratio in which the straight-line 2x + 3y – 20 = 0 divides the join of the points (2, 3) and (2, 10).
Sol:
Given equation is L ≡ 2x + 3y – 20 = 0
L_{11} = 2(2) + 3(3) = 4 + 9– 20 = – 7
L_{22} = 2(2) + 3(10) = 4 + 30 – 20 = 14
We know that the line L = 0 divides the line joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio – L_{11} : L_{22}
= – (– 7) : 14
= 1 : 2
QUESTION24.
State whether (3, 2) (– 4, – 3) are on the same or opposite sides of the straight-line 2x – 3y + 4 = 0.
Sol:
Given equation is L ≡ 2x – 3y + 4 = 0
L_{11} = 2(3) – 3(2) + 4 = 6 – 6 + 4 = 4 > 0
L_{22} = 2(– 4) – 3(–3) + 4 = – 8 + 9 + 4 = 5 > 0
L_{11}, L_{22} has the Same sign
∴ Given points lies the same side of the line 2x – 3y + 4 = 0.
QUESTION25.
Find the ratio in which (i) X-axis and (ii) Y-axis divide the line segment joining A (2, –3) and B (3, – 6)
Sol:
(i) X – axis divide in the ratio – y_{1} : y_{2 } = – ( –3) : – 6 = – 1 : 2
(ii) Y – axis in the ratio – x_{1} : x_{2 = } – 2 : 3 = – 2 : 3
Find the equation of the straight line passing through the point of intersection of the lines x + y + 1 = 0 and 2x – y + 5 = 0 and containing the point (5, – 2).
Sol:
Given equations are
x + y + 1 = 0 ……… (1)
2x – y + 5 = 0……. (2)
The equation of the line passing through the point of intersection of the lines L_{1} = 0 and L_{2} = 0 is L_{1} + λ L_{2} = 0
The equation of the line passing through the point of intersection of the lines (1) and (2) is
x + y + 1 + λ (2x – y + 5) = 0
but the line passes through (5, – 2)
5 – 2 + 1 + λ (2(5) – (– 2) + 5) = 0
4 + λ (10 + 2 + 5) = 0
4 + λ (17) = 0
λ = –4 /17
∴ the equation of required line is
x + y + 1 + (–4 /17) (2x – y + 5) = 0
17(x + y + 1) –4 (2x – y + 5) = 0
17x + 17y + 17 – 8x + 4y – 20 = 0
9x + 21y – 3 = 0
3x + 7y – 1 = 0
QUESTION27.
If a, b, and c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
Sol:
Given, a, b, and c are in A.P. and the equation is ax + by + c = 0
a, b, and c are in A.P.
⟹ b = (a + c)/2
Substitute b value in the equation ax + by + c = 0
⟹ ax + [ (a + c)/2]y + c = 0
ax + ay/2 + cy/2 + c = 0
a(x + y/2) + c(y/2 + 1) = 0
dividing on both sides by a
⟹ (x + y/2) + (c/a) (y/2 + 1) = 0
It is in the form of L_{1} + λ L_{2} = 0
∴ given equation represent a family of concurrent lines
x + y/2 = 0 and y/2 + 1 = 0 are the lines
y/2 + 1 = 0 ⟹ y/2 = – 1
y =– 2 ⟹ x = 1
The point of concurrency is (– 2, 1)
QUESTION28.
Find the point of intersection of the line 7x + y + 3 = 0 and x + y = 0
Sol:
Given equations are 7x + y + 3 = 0 and x + y = 0
∴ The point of intersection is
Transform the following equations into the form L_{1} + λ L_{2} = 0 and find the point of concurrency of the family of straight lines represented by the equation.
(i) (2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0
Given equation is
(2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0
2x + 5kx – 3y – 6ky + 2 – k = 0
2x – 3y + 2 + k (5x – 6y – 1) = 0
It is in the form of L_{1} + λ L_{2} = 0 L_{1 }≡ 2x – 3y + 2 = 0 and L_{2}≡ 5x – 6y – 1 = 0
By solving above equations, we get point of concurrency
∴ The point of concurrency is (5, 4)
(ii) ( k + 1) x + (k + 2) y + 5 = 0
Given equation is (k + 1) x + (k + 2) y + 5 = 0
kx + x + ky + 2y + 5 = 0
x + 2y + 5 + k (x + y) = 0
It is in the form of L_{1} + λ L_{2} = 0
L_{1 }≡ x + 2y + 5 = 0 and L_{2}≡ x + y = 0
By solving the above equations, we get point of concurrency
∴ The point of concurrency is (5, – 5)
QUESTION30.
Find the value of ‘p’, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.
Sol:
Given that x + p = 0 —— (1), y + 2 = 0 —— (2) and 3x + 2y + 5 = 0 —— (3) are concurrent.
From (1) x + p = 0 ⟹ x = – p
From (2) y + 2 = 0 ⟹ y = – 2
Point of intersection of (1) and (2) is (– p, – 2)
From (3)
3y + 2x + 5 = 0
3(– p) + 2 (– 2) + 5 = 0
– 3p – 4 + 5 = 0
– 3p + 1 = 0
3p = 1
p = 1/3
Find the area of the triangle formed by the following straight lines and the coordinate axes.
(i) x – 4y + 2 = 0
The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes is
The area of the triangle formed by the straight-line x – 4y + 2 = 0 with the coordinate axes is =
(ii) 3x – 4y + 12 = 0
The area of the triangle formed by the straight-line 3x – 4y + 12 = 0 with the coordinate axes is =
QUESTION32.
Find the angle between the lines 2x + y + 4 = 0 and y – 3x =7
Sol:
Given Equations are 2x + y + 4 = 0 and y – 3x =7
If θ is the angle between the lines a_{1}x + b_{1}y + c = 0 and a_{2}x + b_{2}y + c = 0, then cos θ =
∴ θ = 45^{0}
QUESTION33.
Find the perpendicular distance from the point (– 3, 4) to the straight line 5x – 12y = 2
Sol:
Given equation is 5x – 12y = 2 ⟹ 5x – 12y – 2 = 0
The perpendicular distance from the point (x_{1}, y_{1}) to the straight-line ax + by + c = 0 is
The perpendicular distance from the point (– 3, 4) to the straight-line 5x – 12y – 2 = 0 is
= 5
QUESTION34.
Find the distance between the parallel lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0
Sol:
Given equations are 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0
3x + 4y – 3 = 0 ⟹ 2(3x + 4y – 3) = 2(0) (multiplying on both sides by 2)
⟹ 6x + 8y – 6 = 0
The distance between two parallel lines ax + by + c_{1} =0 and ax + by + c_{2} =0 is
The distance between two parallel lines 6x + 8y – 6 = 0 and 6x + 8y – 1 = 0 is
Find the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0
Sol:
Let (h, k) be the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0
If (h, k) is the foot of the perpendicular from (x_{1}, y_{1}) on the straight-line ax + by + c = 0, then
h+ 1 = 5 and k – 3 = – 1
h = 4 and k = 2
∴ (h, k) = (4, 2)
QUESTION36.
Find the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0
Sol:
Let (h, k) be the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0
If (h, k) is the image of the point (x_{1}, y_{1}) with respect to the straight line ax + by + c = 0, then
If (h, k) is the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0,
h – 1 = – 4 and k + 2 = 6
h =– 3 and k = 4
∴ (h, k) = (– 3, 4)
QUESTION37.
If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β), find α + β.
Sol:
Given 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β)
⟹ (α, β) is the image of (3, –4) concerning the straight line 2x – 3y – 5 = 0
α – 3 = – 4 and β + 4 = 6
α = – 1 and β = 2
α + β = – 1 + 2 = 1
QUESTION38.
Find the incenter of the triangle whose vertices are (1, ), (2, 0) and (0, 0)
Sol:
Let O = (0, 0), A = (1, ) and B = (2, 0)
in centre of a triangle with sides a, b and c, whose vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is
= =
QUESTION39.
Find the ortho centre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0 and 2x + y – 7 = 0.
Sol:
Given equations are
x + y + 10 = 0 …… (1)
x – y – 2 = 0 …… (2)
2x + y – 7 = 0. …… (3)
Slope of the line (1) is – 1
Slope of the line (2) is 1
The straight lines (1) and (2) are perpendicular lines
The point of intersection of (1) and (2) is the orthocenter
x + y = – 10 ⟹ y = – 10 – x
from (2) x – y – 2 = 0
x – (– 10 – x) – 2 = 0
x + 10 + x – 2 = 0
2x + 8 = 0 ⟹ x = – 4
⟹ y = – 10 + 4 = – 6
∴ Orthocenter is (– 4, – 6)
QUESTION40.
Find the circum centre of the triangle whose sides are x = 1, y = 1 and x + y = 1
Sol:
Given equations are x = 1, y = 1 and x + y = 1
A = (1, 0), B = (1, 1) and C = (0, 1)
AB and BC are perpendicular lines
Circum centre is the midpoint of hypotenuse
i.e., midpoint of AC
The post Straight Lines (2m Questions & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>
Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| Read More »
The post Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| first appeared on Basics In Maths.]]>QUESTION 1
Find the Domain of the following real-valued functions.
It is defined when 6x – x^{2} – 5 ≠ 0
⇒ x^{2} – 6x + 5 ≠ 0
x^{2} – 5x – x + 5 ≠ 0
x (x – 5) –1(x – 5) ≠ 0
(x – 5) (x – 1) ≠ 0
x ≠ 5 or x ≠ 1
∴ domain = R – {1, 5}
It is defined when 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0
⇒ x ≥ –3, x ≤ 3 and x ≠ 0
⇒ –3≤ x, x ≤ 3 and x ≠ 0
⇒ –3≤ x ≤ 3 and x ≠ 0
x ∈ [–3, 3] – {0}
∴ domain = [–3, 3] – {0}
It is defined when x + 2 ≥ 0, 1 – x > 0 and 1 – x ≠ 0
⇒ x ≥ –2, x < 1 and x ≠ 0
x ∈ [–2, ∞) ∩ (– ∞, 1) – {0}
⇒ x ∈ [–2, 1) – {0}
∴ domain = [–2, 1) – {0}
It is defined when 4x – x^{2} ≥ 0
⇒ x^{2} – 4x ≤ 0
x (x – 4) ≤ 0
(x – 0) (x – 4) ≤ 0
x ∈ [0, 4]
∴ domain = [0, 4]
(v) f(x) = log (x^{2} – 4x + 3)
Given f(x) = log (x^{2} – 4x + 3)
It is defined when x^{2} – 4x + 3 > 0
⇒ x^{2 }– 3x – x + 3 > 0
x (x – 3) –1(x – 3) > 0
(x – 3) (x – 1) > 0
x ∈ (–∞, 1) ∪ (3, ∞)
x ∈ R – [1, 3]
∴ domain = R – [1, 3]
Given f(x) = It is defined when x^{2} – 1 ≥ 0 and x^{2} – 3x + 2 > 0
(x + 1)(x – 1) ≥ 0 and x^{2} – 2x – x + 2 > 0
(x + 1) (x – 1) ≥ 0 and x (x – 2) (x – 1) > 0
x∈ (–∞, –1) ∪ (1, ∞) and x ∈ (–∞, –1) ∪ (2, ∞)
∴ domain = R – (–1, 2]
⇒ x < 0
∴ domain = (–∞, 0)
⇒ x > 0
∴ domain = (0, ∞)
QUESTION 2
If f : R→ R , g : R → R defined by f (x ) = 4x – 1, g(x) = x^{2} + 2 then find (i) (gof) (x) (ii) (gof) () (iii) (fof) (x) (iv) go(fof) (0).
Sol: Given f(x) = 4x – 1, g(x) = x^{2} + 2
(i) (gof) (x) = g (f (x))
= g (4x – 1)
= (4x – 1)^{2} + 2
= 16x^{2} – 8x + 1 + 2
= 16x^{2} – 8x + 3
= g (a + 1 – 1)
= g(a)
= a^{2} + 2
(iii) (fof) (x) = f (f (x))
= f (4x – 1)
= 4 (4x – 1) – 1
= 16x – 4 – 1
= 16x – 5
(iv) go(fof) (0) = g(fof) (0)
= g (f (f (0)))
= g (f (– 1))
= g (– 4 – 1)
= g (– 5)
= (– 5)^{2} + 2
= 25 + 2 = 27
QUESTION 3
If f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x^{2}, then find (i) (3f – 2g) (x) (ii) (fg)(x) (iii) (x) (iv) (f + g+ 2) (x)
Sol: Given f(x) = 2x – 1 and g(x) = x^{2}
(i) (3f – 2g) (x) = 3f(x) – 2 g(x)
= 3(2x – 1) – 2(x^{2})
= 6x – 3 – 2x^{2}
(ii) (fg)(x) = f(x) g(x)
= (2x – 1) (x^{2})
= 2x^{3} – 3x^{2}
(iv) (f + g+ 2) (x) = f(x) + g(x) + 2
= 2x – 1 + x^{2} + 2
= x^{2} + 2x + 1
QUESTION 4
If f = {(4, 5), (5, 6), (6, – 4)} g = {(4, – 4), (6, 5), (8,5)}, then find (i) f + g (ii) f – g (iii) 2f +4g (iv) f + 4 (v) fg (vi) f/g (vii) (viii) (ix) f^{2} (x) f^{3}
Sol: Given f = {(4, 5), (5, 6), (6, – 4)}
g = {(4, – 4), (6, 5), (8,5)}
The domain of f ∩ The Domain of g = {4, 6}
(i) (f + g) (4) = f (4) + g (4)
= 5 – 4 = 1
(f + g) (6) = f (6) + g (6)
=– 4 + 5 = 1
∴ f + g = {(4, 1), (6, 1)}
(ii) (f – g) (4) = f (4) – g (4)
= 5 – (– 4) = 5 + 4 = 9
(f – g) (6) = f (6) – g (6)
= – 4– 5 = – 9
∴ f – g = {(4, 9), (6, – 9}
(iii) (2f +4g) (4) = 2 f (4) + 2 g (4)
= 2(5) + 4 (– 4)
= 10 – 16
=– 6
(2f +4g) (6) = 2 f (6) + 2 g (6)
= 2(– 4) + 4 (5)
= – 8 + 20
=12
∴ (2f +4g) = {(4, – 6), (6, 12)}
(iv) (f + 4) (4) = f (4) + 4 = 5 + 4 = 9
(f + 4) (5) = f (5) + 4 = 6 + 4 = 10
(f + 4) (6) = f (6) + 4 = – 4 + 4 = 0
∴ (f + 4) = {(4, 9), (5, 10), (6, 0)}
(v) fg (4) = f (4) g (4) = (5) (– 4) =– 20
fg (6) = f (6) g (6) = (– 4) (5) =– 20
∴ fg = {(4, – 20), (6, – 20)}
(vi) f/g (4) = f(4)/g(4) = 5/ – 4 = – 5/4
f/g (6) = f(6)/g(6) = – 4/ 5
∴ f/g = {(4, – 5/4), (6, – 4/ 5)}
(ix) f^{2}(4) = (f (4))^{2} = (5)^{2} = 25
f^{2}(5) = (f (5))^{2} = (6)^{2} = 36
f^{2}(6) = (f (6))^{2} = (– 4)^{2} = 16
∴ f^{2} = {(4, 25), (5, 36), (6, 16)}
(x) f^{3}(4) = (f (4))^{3} = (5)^{3} = 125
f^{3}(5) = (f (5))^{3} = (6)^{3} = 216
f^{3}(6) = (f (6))^{3} = (– 4)^{3} = –64
∴ f^{3} = {(4, 125), (5, 216), (6, –64)}
QUESTION 5
If A = {0, π/6, π/4, π/3, π/2} and f: A→ B is a surjection defined by f(x) = cos x, then find B.
Sol: Given A = {0, π/6, π/4, π/3, π/2}
f(x) = cos x
f (0) = cos (0) = 1
f(π/3) = cos (π/3) = 1/2
f(π/2) = cos (π/2) = 0
QUESTION 6
If A = {–2, –1, 0, 1, 2} and f: A→ B is a surjection defined by f(x) =x^{2} + x + 1, then find B.
Sol: Given A = {–2, –1, 0, 1, 2} and f(x) = x^{2} + x + 1
f (–2) = (–2)^{2} + (–2) + 1= 4 – 2 + 1 = 3
f (–1) = (–1)^{2} + (–1) + 1= 1 – 1 + 1 = 1
f (0) = (0)^{2} + (0) + 1= 0 + 0 + 1 = 1
f (1) = (1)^{2} + (1) + 1= 1 + 1 + 1 = 3
f (2) = (2)^{2} + (2) + 1= 4 + 2 + 1 = 7
∴ B = {1, 3, 7}
QUESTION 7
If A = {1, 2, 3, 4} and f: A→ B is a surjection defined by f(x) = then find B.
Sol: Given A = {1, 2, 3, 4} and f(x) =
QUESTION 8
If f(x) = 2, g(x) = x^{2}, h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x)
Sol: Given f(x) = 2, g(x) = x^{2}, h(x) = 2x
(fo(goh)) (x) = fo (g (h (x))
= f (g (h (x))
= f(g(2x)
= f((2x)^{2})
= f(4x^{2}) = 2
QUESTION 9
If f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x^{2} +1 then find (i) (gof^{-1}) (2) (ii) (gof) (x – 1)
Sol: Given f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x^{2} +1
(i) Let y = f(x) ⟹ x = f^{-1}(y)
y = 3x – 2
y + 2 = 3x
x = (y + 2)/3
f^{-1}(y) = (y + 2)/3
∴ f^{-1}(x) = (x + 2)/3
Now
(gof^{-1}) (2) = g(f^{-1}(2))
= g ((2 + 2)/3)
= g (4/3)
= (4/3)^{2} + 1 = 16/9 + 1 = 25/9
(ii) (gof) (x – 1) = g (f (x – 1))
= g [ 3(x – 1) – 2)]
= g (3x – 3 – 2)
=g (3x – 5)
= (3x – 5)^{2} + 1
= 9x^{2} – 30x + 25 + 1
= 9x^{2} – 30x + 26
QUESTION 10
If f: N→ N defined by f (x) = 2x + 5, Is onto? Explain with reason.
Sol: Given f (x) = 2x + 5
Let y = f(x) ⟹ x = f^{-1}(y)
y = 2x + 5
2x = y – 5
x = (y – 5)/2 ∉ N
∴ f(x) is not onto
QUESTION 11
Find the inverse of the following functions
(i) If a, b ∈ R, f: R→ R defined by f(x) = ax + b
Given function is f(x) = ax + b
Let y = f(x) ⟹ x = f^{-1}(y)
y = ax + b
y – b = ax
x = (y – b)/a
f^{-1}(y) = (y – b)/a
∴ f^{-1}(x) = (x – b)/a
(ii) f: R→ (0, ∞) defined by f(x) = 5^{x}
Given function is f(x) = 5^{x}
Let y = f(x) ⟹ x = f^{-1}(y)
y = 5^{x}
(iii) f: (0, ∞) → R defined by f(x) =
Let y = f(x) ⟹ x = f^{-1}(y)
x = 2^{y}
f^{-1}(y) = 2^{y}
∴ f^{-1}(x) = 2^{x}
(iv) f: R→ R defined by f(x) = e^{4x + 7}
Given function is f(x) = e^{4x + 7}
Let y = f(x) ⟹ x = f^{-1}(y)
y = e^{4x + 7}
(v) f: R→ R defined by f(x) = (2x + 1)/3
Given function is f(x) = (2x + 1)/3
Let y = f(x) ⟹ x = f^{-1}(y)
y= (2x + 1)/3
3y = 2x + 1
2x = 3y – 1
x = (3y – 1)/2
f^{-1}(y) = (3y – 1)/2
∴ f^{-1}(x) = (3x – 1)/2
QUESTION 12
If f: R→ R defined by f(x) = , then show that f (tan θ) = cos 2θ
∴ f (tan θ) = cos 2θ
QUESTION 13
If f: R – {±1} → R defined by f(x) = , then show that = 2f (x)
QUESTION 14
If the function f: R→ R defined by f(x) = , then show that f (x + y) + f (x – y) = 2 f(x) f(y).
∴ f (x + y) + f (x – y) = 2 f(x) f(y)
QUESTION 15
If f(x) = cos (log x), then show that = 0
Sol: Given function is f(x) = cos (log x)
= cos (log x) cos (log y) – cos (log x) cos (log y)
= 0
Hence proved
QUESTION 16
Find the range of the following real-valued functions
Sol:
e^{y} > 0
∴ Range of f(x) is R
It is defined when x – 2 ≠ 0
⟹ x ≠ 2
Domain = R – {2}
y = x + 2
if x = 2 ⟹ y = 4
∴ Range of f(x) is R – {4}
QUESTION 17
Find the domain and range of the following real valued functions
Sol:
It is defined when 1 + x^{2} ≠ 0
⟹ x^{2} ≠ – 1
x ∈ R
∴ domain of f(x) is R
y (1 + x^{2}) = x
y + x^{2}y = x
x^{2} y – x + y = 0
It is defined when 1 – 4y^{2} ≥ 0 and 2y ≠ 0
⟹ 4y^{2} – 1 ≤ 0 and y ≠ 0
(2y – 1) (2y + 1) ≤ 0 and y ≠ 0
(y – 1/2) (y + 1/2) ≤ 0 and y ≠ 0
– ½≤ y ≤ ½ and y ≠ 0
∴ Range of f(x) is [– ½, ½] – {0}
It is defined when 9 – x^{2 }≥ 0
⟹ x^{2} – 9 ≤ 0
(x + 3) (x – 3) ≤ 0
– 3 ≤ y ≤ 3
∴ the domain of f (x) is [– 3, 3]
y^{2} = 9 – x^{2}
x^{2} = 9 – y^{2 }
It is defined when 9 – y^{2 }≥ 0
⟹ y^{2} – 9 ≤ 0
(y + 3) (y – 3) ≤ 0
– 3 ≤ y ≤ 3
y ∈ [– 3, 3]
∴ Range of f (x) is [0, 3] (∵ y ≥ 0)
clearly, x ∈ R
∴ domain of f(x) is R
If x = 0, then y = 1
If x = – 1, then y = 1
If x = 1, then y = 3
If x = – 2, then y = 3
If x = 2, then y = 5
∴ Range of f (x) is [1, ∞)
(iv) f(x) = [x]
clearly Domain = R and Range = Z
QUESTION 18.
Define (i) One – One function (ii) Onto function (iii) Bijection (iv) Even and Odd functions
(i) One – One Function: one – one, if every element of A has a unique image in B.
(ii) Onto function: A function f: A→ B is said to be onto if ∀ y ∈ B there exists x ∈ A such that f(x) = y.
(iii) Bijection: A function f: A→ B is said to be Bijection if it is both one-one and onto.
(iv) Even and Odd functions:
If f(–x) = f(x), then f(x) is even function
If f(– x)
= – f(x), then f(x) is odd function
The post Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| first appeared on Basics In Maths.]]>
Natural numbers: All the counting numbers starting from 1 are called Natural numbers.
1, 2, 3… Etc.
Whole numbers: Whole numbers are the collection of natural numbers including zero.
0, 1, 2, 3 …
Integers: integers are the collection of whole numbers and negative numbers.
….,-3, -2, -1, 0, 1, 2, 3,…..
Integers on a number line:
Operations on integers:
3 + 4 = 7
-2 + 4 = 2
Subtraction of integers on a number line:-
6 – 3 = 3
Multiplication of integers on a number line:-
2 × 3 ( 2 times of 3) = 6
3 × (- 4 ) ( 3 times of -4) = -12
Multiplication of two negative integers:
Ex:- -3 × -2 = 6, -10 × -2 = 20 and so on.
Multiplication of more than two negative integers:
• If we multiply three negative integers, then the result will be a negative integer.
Ex:- -3 × -4 × -5 = -60, -1× -7 × -4 = -28 and so on.
• If we multiply four negative integers, then the result will be a positive integer.
Ex:- -3 × -4 × -5 × -2 = 120, -1× -7 × -4 × -2 = 56 and so on.
Note:-
1. If the no. of negative integers is even, then the result will be positive.
2. If the no. of negative integers is odd, then the result will be negative.
Division of integers:
Ex:- -3 ÷ 1 = 3, 4 ÷ -2 = -2 and so on.
Ex:- -3 ÷ -1 = 3, -4 ÷ -2 = 2 and so on.
Properties of integers:
1.Closure property:-
2.commutative property:-
3.associative property:-
Additive identity:-
1 + 0 = 0 + 1 = 1, 10 + 0 = 0 + 10 = 10
•For any integer ‘a’, a + 0 = 0 + a
•0 is the additive identity.
Additive inverse:-
2 + (-2) = (-2) + 2 = 0, 5 + (-5) = (-5) + 5 = 0
•For any integer ‘a’, a+ (-a) = (-a) + a = 0
•Additive inverse of a = -a and additive inverse of (-a) = a
Multiplicative identity:-
2 × 1 = 1 × 2 = 2, 5 × 1 = 1 × 5 = 5
•For any integer ‘a’, a × 1 = 1 × a = a
•1 is the multiplicative identity.
multiplicative inverse:-
For any integer ‘a’, 1/a × a = a × 1/a = 1
distributive property:-
For any three integers a, b and c, a × (b + c) = (a × b) + (a × c).
3 × (2 + 4) = 18
(3 × 2) + (3 × 4) = 6 + 12 = 18
∴ 3 × (2 + 4) = (3 × 2) + (3 × 4).
Fraction: A fraction is a number that represents a part of the whole. A group of objects is divided into equal parts, then each part is called a fraction.
The proper and improper fractions:
In a proper fraction, the numerator is less than the denominator.
Ex: – 1/5, 2/3, and so on.
In an improper fraction, the numerator is greater than the denominator.
Ex: – 5/2,11/5 and so on.
Comparing fractions:
Like fractions: – We have to compare the like fractions with the numerator only because the like fractions have the same denominator. The fraction with the greater numerator is greater and the fraction with the smaller numerator is smaller.
Unlike fractions: –
With the same numerator: For comparing unlike fractions, we have to compare denominators when the numerator is the same. The fraction with a greater denominator is smaller and the fraction with a smaller denominator is smaller.
Note: – To find the equivalent fractions of both the fractions with the same denominator, we have to take the LCM of their denominators.
Addition of fractions:
∗ Like Fractions:
∗ Unlike fractions:
Subtraction of fractions:
∗ Like fractions:
Unlike fractions: – First, we have to find the equivalent fraction of given fractions and then subtract them as like fractions
Multiplication of fractions:
Multiplication of fraction by a whole number: –
Multiplication of numbers means adding repeatedly.
• To multiply a whole number with a proper or improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator the same.
2.Multiplication of fraction with a fraction: –
multiplication of two fractions =
Division of fractions:
⇒ 6 one-thirds in two wholes
Reciprocal of fraction: reciprocal of a fraction is .
Note:
1.Division of a whole number by a fraction: –
2.Division of a fraction by another fraction: –
Decimal number or fractional decimal:
In a decimal number, a dot(.) or a decimal point separates the whole part of the number from the fractional part.
The part right side of the decimal point is called the decimal part of the number as it represents a part of 1. The part left to the decimal point is called the integral part of the number.
Note: –
Comparison of decimal numbers:
while comparing decimal numbers, first we compare the integral parts. If the integral parts are the same, then compare the decimal part.
Ex: – which is bigger: 13.5 or 14.5
Ans: 14.5
Which is bigger: 13.53 or 13. 25
Ans: 13.53
Multiplication of decimal numbers:
For example, we multiply 0.1 × 0.1
Multiplication of decimal numbers by 10, 100, and 1000: –
Here, we notice that the decimal point in the product shifts to the right side by as many zeroes as in 10, 100, and 1000.
Division of decimal number:
Division of decimal number by 10,100 and 1000: –
Here, we notice that the decimal point in the product shifts to the left side by as many zeroes as in 10, 100, and 1000.
Rational numbers:
The numbers which are written in the form of p/q, where p, q are integers, and q ≠ 0, are called rational numbers.
Rational numbers are a bigger collection of integers, negative fractional numbers, positive fractional numbers.
Ex: – 1, 2, -1/2, 0 etc.
Equation: Equation is the condition on a variable. It says that two expressions are equal.
Ex: – x + 2 = 5; 2 = x + 3
Balanced equation:
In an equation, if LHS =RHS, then that equation is balanced.
If the same number is added or subtracted on both sides of the balanced equation, the equation remains will the same.
Ex: 8 + 3 = 11
If add 2 on both sides ⇒ LHS = 8 + 3 + 2 = 13
RHS = 11 + 2 = 13
∴ LHS = RHS
8 +3 = 11 if subtract 2 on both sides
LHS = 8 + 3 – 2 = 9
RHS = 11 – 2 = 9
∴ LHS = RHS
Using algebraic equations in solving day to day problems:
Complimentary angles: When the sum of the angles is 90^{0}, the angles are called complementary angles.
Ex: 30^{0}, 60^{0}; 20^{0}, 70^{0} and soon.Supplementary angles: When the sum of the angles is 180^{0}, the angles are called Supplementary angles.
Ex: 120^{0}, 60^{0}; 110^{0}, 70^{0} and soon.
Adjacent angles: The angle having a common Arm and a common vertex are called Adjacent angles.
⇒ ∠AOC and ∠BOC adjacent angles.
Vertically opposite angle: If two lines are intersecting at a point, then the angles that are formed opposite to each other at that point are called vertically opposite angles.
Transversal: A line that intersects two or more lines at distinct points is called a transversal.
Angles made by a transversal:
Corresponding angles: –
Two angles which are lies on the same side of the transversal and one interior and another one exterior are called corresponding angles.
∠1, ∠5; ∠2, ∠6; ∠3, ∠7 and ∠4, ∠8
Alternate angles: –
Two angles which are the lies opposite side of the transversal and both interior or exterior are called corresponding angles.
∠1, ∠7; ∠2, ∠8 are exterior alternate angles
∠3, ∠5; ∠4, ∠6 are interior alternate angles.
∠3, ∠6; ∠4, ∠5 interior angles same side of the transversal.
Transversal on parallel lines:
• If pair of parallel lines are intersected by a transversal then the angles of each pair of corresponding angles are equal
⇒ ∠1, =∠5; ∠2= ∠6; ∠3= ∠7 and ∠4= ∠8
•If pair of parallel lines are intersected by a transversal then the angles of each pair of interior alternate angles are equal.
∠3= ∠5; ∠4= ∠6
•If pair of parallel lines are intersected by a transversal then the angles of each pair of exterior alternate angles are equal.
∠1= ∠7; ∠2= ∠8
•If pair of parallel lines are intersected by a transversal then the angles of each pair of interior angles on the same side of the transversal are supplementary.
∠3+∠6= 180^{0}; ∠4+ ∠5 = 180^{0}
Note:
Triangle: A closed figure formed by three-line segments is called a triangle.
In ∆ABC,
Classification of triangles:
Triangles can be classified according to the properties of their sides and angles.
According to sides:
Based on sides triangles are three types:
According to angles:
Relationship between the sides of a triangle:
The altitude of a triangle:
We can draw three altitudes in a triangle. A perpendicular line drawn from a vertex to its opposite side of a triangle is called the Altitude of the triangle.
Median of a triangle:
In a triangle, a line drawn from the vertex to the mid-point of its opposite side is called the median of the triangle.
Medians of a triangle are concurrent. We can draw three medians in a triangle.
The point of concurrence of medians is called the centroid of the triangle. It is denoted by G
Angle-sum property of a triangle:
Some of the angles in a triangle is 180^{0}
∠A + ∠B + ∠C = 180^{0}
An exterior angle of a triangle:
When one side of the triangle is produced, the angle thus formed is called an exterior triangle.
Exterior angle property:- The exterior angle of a triangle is equal to the sum of two interior opposite angles.
x^{0}+ y^{0} = z^{0}
Ratio: Comparison of two quantities of the same kind is called ‘Ratio.
The ratio is represented by the symbol ‘:’
If the ratio of two quantities ‘a’ and ‘b’ is a : b, then we read this as ‘a is to b’
The quantities ‘a’ and ‘b’ are called terms of the ratio.
Proportion: if two ratios are equal, then they are said to be proportional.
‘a’ is called as first term or antecedent and ‘b’ is called a second term or consequent.
If a: b = c : d, then a, b, c, d are in proportion and ⇒ ad = bc.
The product of means = the product of extremes
Unitary method: The method in which we first find the value of one unit and then the value of the required no. of units is known as the unitary method.
Direct proportion: In two quantities, when one quantity increase(decreases) the other quantity also increases(decreases) then two quantities are in direct proportion.
Percentages:
‘per cent’ means for a hundred or per every hundred. The symbol % is used to denote the percentage.
1% means 1 out of 100, 17% means 17 out of 100.
Profit and Loss:
Selling price = SP; Cost price = CP
If SP > CP, then we get profit
Profit = SP – CP
SP = CP + profit
If SP < CP, then we get a loss
Loss = CP – SP
SP = CP – Loss
Simple interest:
Principle: – The money borrowed or lent out for a certain period is called the Principle.
Interest: – The extra money, for keeping the principle paid by the borrower is called interest.
Amount: – The amount that is paid back is equal to the sum of the borrowed principal and the interest.
Amount = principle + interest
Interest (I) = where R is the rate of interest.
Data: The information which is in the form of numbers or words and helps in taking decisions or drawing conclusions is called data.
Observations: The numerical entries in the data are called observations.
Arithmetic Mean: The average data is also called an Arithmetic mean.
The arithmetic mean always lies between the highest and lowest observations of the data.
When all the values of the data set are increased or decreased by a certain number, the mean also increases or decreases by the same number.
Mode: The most frequently occurring observation in data is called Mode.
If data has two modes, then it is called bimodal data.
Note: If each observation in a data is repeated an equal no. of times, then the data has no mode.
Median: The middlemost observation in data is called the Median.
Arrange given data in ascending or descending order.
If a data has an odd no. of observations, then the middle observation is the median.
If a data has even no. of observations, then the median is the average of middle observations.
Bar graph:
Bar graphs are made up of uniform width which can be drawn horizontally or vertically with equal spacing between them.
The length of each bar tells us the frequency of the particular item.
Ex:
It represents two observations side by side.
Ex:
Pie chart: A circle can be divided into sectors to represent the given data
Ex:
Budget | Amount in rupees |
Food | 1200 |
Education | 800 |
Others | 2000 |
Savings | 5000 |
Total income | 9000 |
Congruent figures: Two figures are said to be congruent if they have the same shape and size.
Congruency of line segments: If two-line segments have the same length, then they are congruent. Conversely, if two-line segments are congruent, then they have the same length.
Congruency of Triangles:
Two triangles are said to be congruent if (i) their corresponding angles are equal (ii) their corresponding sides are equal.
Ex: In ∆ ABC, ∆ DEF
∠A ≅ ∠D; ∠B≅ ∠E; ∠C ≅ ∠F
AB ≅ DE; BC≅ EF; AC ≅ DF
∴∆ABC ≅ ∆DEF
The criterion for congruency of Triangles:
1.Side -Side -Side congruency (SSS): –
If three side of a triangle is equal to the corresponding three sides of another triangle, then the triangles are congruent.
∴∆ABC ≅ ∆DEF
2.Side -Angle -Side congruency (SAS): –
If two sides and the angle included between the two sides of a triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent.
∴ ∆ABC ≅ ∆DEF
3.Angle – Side -Angle congruency (ASA): –
If two angles and included side of a triangle are equal to the corresponding two angles and included side of another triangle, then the triangles are congruent.
∴ ∆ABC ≅ ∆DEF
4.Right angle – Hypotenuse – Side congruence (RHS): –
If the hypotenuse and one side of a right-angled triangle are equal to the corresponding hypotenuse and side of the other right-angled triangle, then the triangles are Equal.
∴∆ABC ≅ ∆DEF
The no. of measurements required to construct a triangle = 3
A triangle can be drawn in any of the situations given below:
Construction of a triangle when measurements of the three sides are given:
Ex: construct a triangle ABC with sides AB = 4cm, BC = 7cm and AC = 5cm
Step of constructions:
Step -1: Draw a rough sketch of the triangle and label it with the given measurements.
Step -2: Draw a line segment of BC of length 7cm.
Step -3: with centre B, draw an arc of radius 4cm, draw another arc from C with radius 5cm such that it intersects first at A.
Step -4: join A, B and A, C. The required triangle ABC is constructed.
Construction of a triangle when two sides and the included angle given:
EX: construct a triangle ABC with sides AB = 4cm, BC = 6cm and ∠B=60^{0}
Step of constructions:
Step -1: Draw a rough sketch of the triangle and label it with the given measurements.
Step -2: Draw a line segment of AB of length 4cm.
Step -3: draw a ray BX making an angle 60^{0} with AB.
Step -4: draw an arc of radius 5cm from B, which cuts ray BX at C.
Step -5: join C and A, we get the required ∆ABC.
Construction of a triangle when two angles and the side between the angles given:
Ex: construct a triangle PQR with sides QR = 4cm, ∠Q= 120^{0} and ∠R= 40^{0}
Step of constructions:
Step -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a line segment QR of length 4 cm.
Step -3: Draw a ray RX, making an angle 40^{0} with QR.
Step -4: Draw a ray QY, making an angle 100^{0} with QR, which intersects ray RX.
Step -5: Mark the intersecting point of the two rays as P. Required triangle PQR is constructed.
Construction of a triangle when two sides and the non-included angles are given:
Ex: construct a triangle MAN with sides MN = 4cm, AM = 3cm and ∠A= 40^{0}
Step of constructions:
Step -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a line segment MA of length 0f 5cm.
Step -3: Draw a ray AX making an angle 40^{0} with the line segment MA.
Step -4: With M as the centre and radius 3 cm draw an arc to cut ray AX. Mark the intersecting point as N.
Step -5: join M, N, then we get the required triangle MAN.
Construction of a right-angled triangle when hypotenuse and sides are given:
Ex: construct a triangle ABC, right angle at B and AB = 4cm, Ac = 5cm
Step of constructions:
Step -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a line segment BC of length 0f 4cm.
Step -3: Draw a ray BX perpendicular to BC at B
Step -4: Draw an arc from C with a radius of 5cm to intersect ray BX at A.
Step -5: Join A, C, then we get the required triangle ABC.
Variable: It is a dependent term. It takes different value.
Ex: m, x, a, etc.
Constant: It is an independent term. It has a fixed value.
Like terms and Unlike terms: If the terms contain the same variable with the same exponents, then they are like terms otherwise, unlike terms.
Ex: 3x, –4x, x are like terms
3x, 4y, 4 are unlike terms
Coefficient: Coefficient is a term which the multiple of another term (s)
EX: In 5x. 5 is the coefficient of x and x is the coefficient of 5
5 is a numerical coefficient
x is the literal coefficient
Expression: An expression is a single term or a combination of terms connected by the symbols ‘+’ (plus) or ‘−’ (minus).
Ex: 2x – 3. 3x, 2 +3 – 4 etc.
Numerical Expressions: If every term of an expression is constant, then the expression is called numerical expression.
Ex: 2 + 3 + 5, 2 – 4 – 7, 1 + 5 – 4 etc.
Algebraic expression: If an expression at least one algebraic term, then the expression is called an algebraic expression.
Ex: x + y, xy, x – 3, 4x + 2 etc.
Note: Plus (+) and Minus (−) separate the terms
Multiplication (×) and Division (÷) do not separate the terms.
Types of Algebraic expressions:
Monomial: – If an expression has only one term, then it is called a monomial.
Ex: 2x^{2}, 3y, x, y, xyz etc.
Binomial: If an expression has two unlike terms, then it is called binomial.
Ex: 2x+ 3y, x^{2}+ y, x +yz^{2} etc.
Trinomial: If an expression has three unlike terms, then it is called trinomial. Ex: 2x+ 3y + 4xy, x^{2}+ y + z, x^{2} y +yz^{2} + xy^{2} etc.
Multinomial: If an expression has more than three unlike terms, then it is called multinomial.
Ex: 2x+ 3y + 4xy +5, x^{2}+ y + z – 4y + 6 ,
x^{2} y +yz^{2} + xy^{2} – 4xy + 8yz etc.
Degree of a monomial: The sum of all exponents of the variables present in a monomial is called the degree of the monomial.
Ex: Degree of 5xy^{3 }
An exponent of x is 1 and an exponent of y is 3
Sum of exponents = 1 + 3 = 4
∴ degree of 5xy^{3} is 4
Degree of an Algebraic Expression: The highest exponent of all the terms of an expression is called the degree of an Algebraic expression.
Ex: degree of x^{2} + 3x + 4x^{3 }is 3
degree of 3xy + 6x^{2}y + 5x^{2}y^{2} is 4
Addition of like terms:
The sum of two or more like terms is a like term with a numerical coefficient that is equal to the sum of the numerical coefficients of all the like terms in addition.
Ex: 3x + 2x = (3 + 2) x = 5x
4x^{2}y + x^{2}y = (4 + 1) x^{2}y = 5x^{2}y
Subtraction of like terms:
The difference of two like terms is a like term with a numerical coefficient is equal to the difference between the numerical coefficients of the two like terms.
Ex: 3x − 2x = (3 − 2) x = x
4x^{2}y −2 x^{2}y = (4 −2) x^{2}y = 2x^{2}y
Note: (i) addition and subtraction are not done for unlike terms. (ii) If no terms of an expression are alike then it is said to be in the simplified form.
The standard form of an Expression:
In an expression, if the terms are in such a way that the degree of the terms is in descending order, then the expression is said to be in standard form.
Ex: 5 – 2x^{2} + 4x +3x^{3}
Standard form is 3x^{3} – 2x^{2} + 4x + 5
Finding the value of an expression:
Example: find the value of expression x^{3} + y + 3, when x = 1 and y = 2
Sol: given expression is x^{3} + y + 3
Substitute x = 1 and y = 2 in above expression
(1)^{3} + 2 + 3 = 1 + 2 + 3 = 6
Addition of algebraic expressions:
The addition of expressions can be obtained by adding like terms.
This is in two ways: (i) Column or Vertical method (ii) Row or Horizontal method.
Column or Vertical method:
Step –1: Write the expression in standard form if necessary.
Step –2: write one expression below the other such that the like terms come in the same column.
Step –3: Add the like terms column-wise and write the result just below the concerned column.
Ex: Add x^{2} + 3x + 5, 3 – 2x + 3x^{2} and 3x – 2
Row or Horizontal method.
Step –1: Write the expression in standard form if necessary.
Step –2: Re-arrange them term by grouping the like terms.
Step –3: Simplify the coefficients.
Step –4: Write the resultant expression in standard form.
Ex: Add x^{2} + 3x + 5, 3 – 2x + 3x^{2} and 3x –2
Sol: (x^{2} + 3x + 5) + (3 – 2x + 3x^{2}) + (3x –2)
= (x^{2 }+ 3x^{2}) + (3x – 2x + 3x) + (5 + 3 – 2)
= (1 + 3) x^{2} + (3 – 2 + 3) x + 6
=4x^{2} + 4x + 6
Additive inverse of an expression:
For every algebraic expression there exist another algebraic expression such that their sum is zero. These two expressions are called the additive inverse of each other.
Subtraction of algebraic expressions:
This is in two ways: (i) Column or Vertical method (ii) Row or Horizontal method.
Column or Vertical method:
Step –1: Write the expression in standard form if necessary.
Step –2: write one expression below the other such that the expression to be subtracted comes in the second row and the like terms come one below the other.
Step –3: Change the sign of every term of the expression in the second row to get the additive inverse of the expression.
Step –4: Add the like terms column-wise and write the result just below the concerned column.
Ex: Subtract: x^{2} + 3x + 5 from 3x^{2} + 4x – 3
Sol:
Row or Horizontal method:
Step –1: Write the expressions in one row with the expression to be subtracted in a bracket with assigning a negative sign to it.
Step –2: Add the additive inverse of the second expression to the first expression.
Step –3: Group the like terms and add or subtract.
Step –4: Write the resultant expression in standard form.
Ex: Subtract: x^{2} + 3x + 5 from 3x^{2} + 4x – 3
Sol: 3x^{2} + 4x – 3 – (x^{2} + 3x + 5)
= 3x^{2} + 4x – 3 – x^{2} – 3x – 5
= (3 – 1) x^{2} + (4 – 3) x + (– 3 – 5)
= 2x^{2} + x – 8
We know that,
a × a = a^{2} (a raised to the power of 2)
a × a × a = a^{3} (a raised to the power of 3)
a × a × a × a × a × a ×…. m times = a^{m}
a^{m} is in exponential form
a is called base, m is called exponent or index.
Laws of exponents:
Standard form: A number that is expressed as the product of the largest integer exponent of 10 and a decimal number between 1 and 10 is said to be in standard form.
Ex: 1324 in standard form is 1.324 × 10^{3}.
Quadrilateral: A Quadrilateral is a closed figure with four sides, four angles and four vertices.
In Quadrilateral ABCD
Diagonal of a Quadrilateral:
The line segment joining the opposite vertices of a quadrilateral are called the diagonals of the Quadrilateral. In the above figure AC, BD is the diagonals.
Adjacent sides of a Quadrilateral:
The two sides of a Quadrilateral that have a common vertex are called the adjacent sides of the Quadrilateral. From the above figure, AB, BC; BC, CD; CD, DA and DA, AB are the adjacent sides.
Adjacent angles of a Quadrilateral:
The two angles of a Quadrilateral that have a common side are called the adjacent angles of the Quadrilateral. From the above figure, ∠A, ∠B; ∠B, ∠C; ∠C, ∠D and ∠D, ∠A is the adjacent angles.
Opposite sides of a Quadrilateral:
The two sides of a quadrilateral, which do not have a common vertex are called opposite sides of a quadrilateral. From the above figure, AB, CD; BC, DA are the opposite sides.
Opposite angles of a Quadrilateral:
The two angles of a quadrilateral, which do not have a common side are called opposite angles of a quadrilateral. From the above figure, ∠A, ∠C; ∠B, ∠D are the opposite angles.
Interior and exterior of a Quadrilateral:
In a Quadrilateral ABCD, S, N are interior points, M, P are exterior points and A, B, C, D and Q are lies on the Quadrilateral.
Convex Quadrilateral:
A Quadrilateral is said to be a convex Quadrilateral if all line segments joining points in the interior of the Quadrilateral also lie in the interior of the Quadrilateral.
Concave Quadrilateral:
A Quadrilateral is said to be a concave Quadrilateral if all line segments joining points in the interior of the Quadrilateral not lie in the interior of the Quadrilateral.
Angle sum property of a quadrilateral:
The Sum of the angle in a Quadrilateral is 360^{0}
In a Quadrilateral ABCD, ∠A + ∠B + ∠C + ∠D = 360^{0}
Types of Quadrilaterals:
1.Trapezium:
In a Quadrilateral, one pair of opposite sides are parallel then it is Trapezium.
In a Trapezium ABCD, AB∥ DC; AC, BD are diagonals.
2.Kite:
In a Quadrilateral two distinct consecutive pairs of sides are equal in length then it is called a Kite.
In a Kite ABCD, AB = BC; AD = DC AC, BD are diagonals.
3.Parallelogram:
In a Quadrilateral, two pairs of opposite sides are parallel then it is Parallelogram.
In a Parallelogram ABCD, AB∥ DC, AD∥ BC; AD, BD are diagonals.
Properties of parallelogram: –
4.Rhombus:
In a parallelogram in which two adjacent sides are equal, then it is a Rhombus.
In a Parallelogram ABCD, AB∥ DC, AD∥ BC; AD, BD are diagonals.
Properties of Rhombus: –
5.Rectangle:
In a parallelogram all angles are equal, then it is a Rectangle.
Properties of Rectangle: –
6.Square:
In a rectangle adjacent sides are equal, then it is a Square.
Properties of Square: –
Area of a parallelogram:
Area of parallelogram (A) = b × h square units.
The area of the parallelogram is equal to the product of its base (b) and the height(h)
Area of a Triangle:
Area of triangle = ½ b × h square units.
The area of the triangle is equal to half the product of its base (b) and height (h).
In a Right-angled triangle, two of its sides can be the height.
Area of a Rhombus:
The area of the Rhombus is equal to half the product of its diagonals
Area of rhombus = ½ d_{1} × d_{2} square units.
Circumference of the circle:
Circumference of circle = 2πr = πd
Area of the rectangular path:
Area of Rectangular path = area of the outer rectangle – are of the inner rectangle
Net: Net is a short of skeleton-outline in 2d, which when folded the result in 3d shape.
Nets of 3D shapes:
1.Cube:
2.Cylinder:
3.Pyramid:
Oblique Sketches:
Oblique sketches are drawn on a grid paper to visualise 3D shapes.
Ex: Draw an oblique sketch of a 3×3×3 cube
Step-1: Draw the front face
Step-2: Draw the opposite face, which is the same as the front face. The sketch is somewhat offset from Step-1
Step-3: Join the corresponding corners.
Step-4: Redraw using dotted lines for hidden edges.
Isometric Sketches:
Isometric sketches are drawn on a dot isometric paper to visualise 3D shapes.
Ex: Draw an oblique sketch of a 2×3×4 cuboid
Step-1: Draw a rectangle to show the front face.
Step-2: Draw four parallel line segments of length 3cm.
Step-3: Connect the corresponding corners with appropriate line segments
Step-4: This is an isometric sketch of a cuboid
Line of symmetry: The line which divides a figure into two identical parts is called the line of symmetry or axis of symmetry.
An object can have one or more than one lines of symmetry.
Regular polygon:
If a polygon has equal sides and equal angles, then the polygon is called a Regular polygon.
Lines of symmetry for Regular polygons:
Regular polygon | No. of sides | No. of axes of symmetry |
Triangle | 3 | 3 |
Square | 4 | 4 |
Pentagon | 5 | 5 |
Polygon | n | n |
Rotational symmetry: If we rotate a figure, about a fixed point by a certain angle and the figure looks the same as before, then the figure has rotational symmetry.
The angle of rotational symmetry: The minimum angle of rotation of a figure to get the same figure as the original is called the angle of rotational symmetry or angle of rotation.
The angle of rotation of the equilateral triangle is 120^{0}
The angle of rotation of a square is 90^{0}
Order of rotational symmetry:
The no. of times a figure, rotated through its angle of rotational symmetry before it comes to the original position is called the order of rotational symmetry.
The order of rotational symmetry for an equilateral triangle is 3.
The order of rotational symmetry for a square is 4.
Note: All figures have rotational symmetry of order 1, as can be rotated completely through 360^{0} to come back to its original position.
An object has rotational symmetry, only when the order of symmetry is more than 1.
• Some shapes have a line of symmetry and some have rotational symmetry and some have both.
Square, Equilateral triangle and Circle have both line and rotational symmetry.
Visit My Youtube Channel: Clock on Below Logo
The post TS 7th Class Maths Concept first appeared on Basics In Maths.]]>
• Now Compare 5432 and 4678…
5432 is greater as the digits at the ten thousand place in 5432 is greater than that in 4678.
Order of numbers:
arrange the numbers from smallest to the greatest; this order is called Ascending order.
Ex:- 23, 44, 65, 79, 100
arrange the numbers from greatest to the smallest, this order is called Ascending order.
Ex:- 100,79, 65, 33, 23
Formations of numbers
• Form the largest and smallest possible numbers using the digits 3, 2, 4, 1 without repetition
• Largest number formed by arranging the given digits in descending order _ 4321.
• Smallest number formed by arranging the given digits in ascending order _ 1234.
• Greatest two-digit number is 99.
• Greatest three-digit number is 999.
• Greatest four-digit number is 9999.
• Place value is the positional notation, which defines the position of a digit.
Ex:- 3458
8 is one place, 5 is tens place, 4 is hundreds place and 3 is thousands place.
Expanded form
• It refers to expand the numbers to see the value of each digit.
Ex :- 3458 = 3000 + 400 + 50 + 8
= 3×1000 + 4×100 + 5×10 + 8×1
• Note:-
1 hundred = 10 tens
1 thousand = 10 hundreds
1 lakh = 100 thousands = 1000 hundreds
Place value table for Indian system :
Example: Represents the number in 6,35,21,892 in place value table
Place value table for International system :
Ex:- represents the number in 635,218,924 in place value table
• Indian system of numeration:- in the Indian system of numeration we use ones, tens, hundreds, thousands, lakhs and crores. The first comma comes after three digits from the right, the second comma comes two digits latter and the third comma comes after another two digits.E
Ex:- “three crores thirty-five lakh seventeen thousand four hundred thirty” can be written as.3,35,17,430
• International system of numeration:- in the International system of numeration we use ones, tens, hundreds, thousands, millions and billions.
Ex:- “ six hundred thirty-five million two hundred eighteen thousand nine hundred twenty-four” can be written as 635,218,924.
Note:-10 millimetres = 1centimeter
100 centimetres = 1 meter
1000 meters = 1 kilometer
1000 milligrams = 1 gram
1000 grams = 1 kilo gram
Natural numbers: All the counting numbers starting from 1 are called Natural numbers.
1, 2, 3… Etc.
Successor and Predecessor: If we add 1 to any natural number, we get the next number, which is called the Successor. If we subtract 1 from any natural number, we get the previous number, which is called Predecessor.
Ex: – successor of 23 is 24 and predecessor of 32 is 31.
Note:- There is no predecessor of 1 in natural numbers.
Whole numbers: Whole numbers are the collection of natural numbers.
0, 1, 2, 3 …
Representation of whole number on the number line:
• Draw a line mark a point on it.
• Label it as ‘0’
• Mark as many points at equal distance to the right of 0.
• Label the points as 1, 2, 3, 4, … respectively.
• The distance between any two consecutive points is the unit distance.
Addition on the number line:
Addition of the whole number can represent on the number line
Ex:- 3 + 2 = 5
Start from three, we add 3 to 2. We make two jumps to the right of the number line as shown above. We reach at 5.
Subtraction of the whole number can be represented on the number line
Ex :-5 – 3 = 2
Start from 5, we subtract 3 from 5. We make three jumps to the left of the number line shown as above. We reach at 2.
Multiplication on the number line:
For multiplying 2 and 3, start from 0, make 2 jumps using 3 units at a time to the right, as you reach to 6. Thus, 2 × 3 =6.
Properties of whole numbers
Closer property: Two whole numbers are said to be closed if their operation (+, -, ×,÷) is always closed.
Addition:-Whole numbers are closed under addition.
Ex: 3, 2 are whole numbers ⟹ 3 + 2 = 5 ( 5 is whole number)
Subtraction:- Whole numbers are not closed under subtraction as their difference not always a whole number.
Ex:- 2 – 3 = −1 ( −1 is not a whole number)
Multiplication:- Whole numbers are closed under multiplication.
Ex:- 3 × 2 = 6, 6 is a whole number.
Division:- Whole numbers are not closed under division, as their division is not always a whole number.
Ex:- 3 ÷ 2 is not a whole number.
Commutative property: Two whole numbers are said to be commutative if the result is the same when we change their position.
Addition:-Whole numbers are commutative under addition.
Ex: 3, 2 are whole numbers ⟹ 3 + 2 = 5 and 2 + 3 = 5 ( 3 + 2 = 2 + 3).
Subtraction:- Whole numbers are not commutative under subtraction.
Ex:- 2 – 3 = −1 and 3 – 2 = 1( 2 −3 ≠ 3 – 2 ).
Multiplication:- Whole numbers are commutative under multiplication.
Ex:- 3 × 2 = 6 and 2 ×3 = 6 (3 × 2 = 2 ×3)
Division:- Whole numbers are not commutative under division.
Ex:- 3 ÷ 2 ≠ 2 ÷ 3.
Associative property: For any three whole numbers a, b and c if (a ⨀ b)⨀ c = a ⨀ (b ⨀ c), then whole numbers are associative under operation ⨀. [ ⨀ = +, –, × and ÷ ].
Addition:-Whole numbers are associative under addition.
Ex: ( 3 + 2) + 5 = 10 and 3 + (2 + 5) = 10 ⟹ ( 3 + 2) + 5 = 3 + (2 + 5)
Subtraction:- Whole numbers are not associative under subtraction.
Ex:- : ( 3 − 2) − 5 = −4 and 3 − (2 − 5) = 6 ⟹ ( 3 + 2) + 5 ≠3 + (2 + 5)
Multiplication:- Whole numbers are associative under multiplication.
Ex:- (3 × 2) ×5 = 30 and 3 ×(2 × 5) = 30 ⟹ (3 × 2) ×5 = 3 ×(2 × 5)
Division:- Whole numbers are not associative under division.
Ex:- ( 3 ÷ 2) ÷ 5 ≠3 ÷ (2 ÷ 5).
Distributive property:
For any three whole numbers a, b and c, a×(b + c) = (a × b) +( a × c).
Note :Division by zero is not defining.
Identity under addition and multiplication:
2 +0 = 2, 5 + 0 = 5 and so on.
Thus, 0 is the additive identity.
2 ×1 = 2, 4 × 1 = 4 and so on.
Thus, 1 is a multiplicative identity.
Patterns:
The number 3 as shown as
3 can be shown as 6 can be shown as
Divisibility Rule:
The process of checking whether a number is divisible by a given number or not without actual division is called divisibility rule for that number.
Divisibility by 2:- a number is divisible by 2 if its once place is either 0, 2, 4, 6 or 8.
Ex:- 26 is divisible by 2. 35 not divisible by 2.
Divisibility by 3:- if the sum of the digits of a number is divisible by 3, then that number is divisible by 3.
Ex:- 231 → 2 + 3 +1 =6, 6 is divisible by 3
∴ 231 is divisible by 3
436 → 4 + 3 + 6 = 13, 13 is not divisible by 3
∴ 436 is not divisible by 3.
Divisibility by 4:- if the last two digits of a number is divisible by 4, then that number is divisible by 4.
Ex:- 436, 36 is divisible by 4 ∴ 436 is divisible by 4
623, 23 is not divisible by 4 ∴ 623 is not divisible by 4.
Divisibility by 5:- a number is divisible by 5, if its once place is either 0 or 5.
Ex:- 20, 25 are divisible by 5. 22, 46 are not divisible by 5.
Divisibility by 6:- a number is divisible by 6, if it is divisible by both 3 and 2.
Ex:- 242 is divisible by both 2 and 3 ∴ 242 is divisible by 6
232 is divisible by 3 but not 2 ∴ 232 is not divisible by 6
Divisibility by 8:- if the last three digits of a number is divisible by 8, then that number is divisible by 8.
Ex:- 4232, last three digits 232 are divisible by 8
∴ 4232 is divisible by 8.
Divisibility by 9:- if the sum of the digits of a number is divisible by 9, then that number is divisible by 9.
Ex:- 459, 4 + 5 + 9 = 18 → 18 is divisible by 9 ∴ 459 is divisible by 9
532, 5 + 3 + 2 = 10 → 10 is not divisible by 9 ∴ 532 is not divisible by 9.
Divisibility by 10:- a number is divisible by 10 if its once place is 0.
Ex:- 20 is divisible by 10. 22, 45 are not divisible by 10.
Divisibility by 11:- A number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places is either 0 or 11.
Ex:- 6545
Sum of the digits at odd places = 5 + 5 = 10
Sum of the digits at even places = 4 + 6 = 10
Now difference is 10 – 10 = 0
∴ 6545 is divisible by 11.
Factors: a number which divides the other number exactly is called a factor of that number.
6 = 1×6
= 2×3 ⟹ factors of 6 are: 1, 2, 3 and 6
Note- 1)1 is a factor of every number.
2) Every number is a factor of itself.
3) Every factor is less than are equal to the given number.
4) Factors of a given number are countable.
Prime numbers: The numbers, which have only two factors 1, and itself are called prime numbers.
2, 3, 5, 7, …. Are prime numbers
Composite numbers: The number, which has more than two factors are called composite numbers.
4, 6,8,9….. are composite numbers.
2) 2 is the smallest prime number
3) 4 is the smallest composite number.
Co – prime number: The number which has no common factor except 1 is called co-prime number.
Ex:- (2, 3), (4,5) ……
Twin – primes: If the difference of two prime numbers is 2, then those numbers are called twin prime numbers.
Ex:- (2,3), (3,5), (17,19)…..
Factorization: When a number is expressed as the product of its factors, we say that the number has been factorized. The process of finding the factors is called Factorisation.
Ex:- factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24
24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
Prime factorisation: The process of finding the prime factors is called prime factorisation.
Ex:- 24 = 2 × 12
2 × 3 × 4
2 × 3 × 2 × 2
∴ Prime factorisation of 24 is 2 × 2 × 2 × 3.
Methods of prime factorization:
Division method:- Prime factorization of 12 using the division method,
fallow the procedure.
Start dividing by the least prime factor. Continue division till the resulting number to be divided is 1.
The prime factorization of 12 is 2 × 2 × 3.
Factor tree method:- To find the prime factorization of 24, using the factor tree method we proceed as follows:
Common factors: Common factors are those numbers, which are factors of all the given numbers.
Ex:- 12, 9
Factors of 12 are: 1, 2, 3, 4, 6 and 12
Factors of 9 are: 1, 3 and 9
∴ Common factors of 12, 9 are 1,3
Highest Common Factor (H.C.F):- The highest common factor of two or more numbers is the highest of their common factors. It is also called ad Greatest Common Divisor(G.C.D).
Ex:- H.C.F of 12, 9
Factors of 12 = 1, 2, 3,4, 6, 12
Factors of 9 = 1, 3, 9
Common factors of 12, 9 = 1,3
Highest common factor is 3
∴ H.C.F of 12, 9 is 3
The HCF of 9 , 12 can be found by the prime factorization method as follows.
9 = 3 × 3
12 =3 × 2× 2
The common factor of 12, 9 is 3
∴ H.C.F of 12, 9 is 3
Continue division method:
Euclid invented this method. Divide the larger number by smaller and then divide the previous divisor by the remainder until the remainder zero. The last divisor is the HCF of given numbers.
Common multiple multiples of 3 are 3, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39,42…
Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52….
Common multiples of 3 and 4 are 12, 24, 36….
Least common multiple (LCM):- The least common multiple of two or more given numbers is the lowest of their common multiple.
Ex:- LCM of 3 and 4
Multiples of 3 = 3, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39,42…
Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52….
Common multiples of 3 and 4 = 12, 24, 36….
∴ LCM of 3, 4 is 12.
Methods of finding LCM:
1. Prime factorization method:- the LCM of 6, 15 by using prime factorization method is as follows:
i) Express each number as the product of prime factors
Prime factors of 6 = 2 × 3
Prime factors of 15 = 5× 3
ii) Take the common factors both: 3
iii) Take the extra factors of both 6 and 15 i.e., 2 and 5
iv) The product of all common factors of two numbers and extra common factors of both finds LCM.
∴ LCM of 6 and 15 = (3) × 2 × 5 = 30.
2. Division method:- To find LCM of 6 and 15:
i. Arrange the given numbers in a row.
ii. Then divide the least prime number, which divides at least two of the given numbers, and carry forward the numbers, which are not divisible by that number if any.
iii.Repeat the process till no numbers have a common factor other than 1.
iv. LCM is the product of the divisors and the remaining numbers.
Ex:-
∴ LCM of 6, 15 = 3 × 2 × 5 = 30.
Note:- the product of LCM and HCF of given numbers = the product of given numbers.
Ex:- 6 × 15 = 30 × 3 = 90.
The term ‘geometry’ is derived from the Greek word ‘geometron’.
Geo means Earth and metron means measurement.
Point: Point is a location or position on the surface of the plane. It is denoted by capital letters of the English alphabet.
Line: It is made up of infinitely many points with infinity length.
Ray: Ray is a part of a line. It begins at a point and goes on endlessly n a specific direction.
It is denoted by
Line segment: It is a part of the line with the finite length.
Intersecting lines: If two lines are meeting at the same point, then those lines are called intersecting lines. That pint is called the point of intersection.
Parallel lines: The lines, which are never meet at any point, are called parallel lines.
Curve: Anything, which is not straight, is called Curve.
Simple curve: – A curve that does not cross itself.
Open curve: – A curve in which its endpoints do not meet.
Closed curve: – A curve that has no endpoint is called a closed curve.
∗ A closed curve has three parts
The Interior of the curve: – It refers to the inside area of the curve. (B)
The exterior of the curve: – It refers to the outside area of the curve. (A)
On the curve: – It refers to the inside area of the curve. (C)
Polygon: – polygon is a simple closed figure made by line segments.
Angle: the figure formed by two rays having a common end is called an angle.
Here two rays OA, OB are arms of the angle
O is the Vertex. It is denoted by ∠AOB or ∠ BOA.
Triangle: A simple closed figure formed by the three line segments is a triangle. The line segments are called sides of the triangle.
Quadrilateral: A simple closed figure formed by the four-line segments is a Quadrilateral.
Circle: The set of points that are at a constant distance from a fixed point is called a circle. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle.
Circumference of the circle: – the length of the boundary of the circle is called the circumference of the circle.
Arc: – The part of the circumference is called Arc. From the above fig. is an arc.APisarc.
Sector: – Region enclosed by an arc and two radii is called a sector.
Segment: – The region enclosed by arc and chord is called a segment of the circle.
Measure of line segment:
Simple observation: – We can tell which line segment is greater than other just by observing the two-line segments but it is not sure.
Here we can clearly say that CD > AB but sometimes it is difficult to tell which one is greater.
But this is a difficult method because every time to measure the different size of line segments we have to make a separate line segment.
Comparing by instruments: – To compare any two-line segments accurately, we use ruler(scale) and divider.
∗ We can use a ruler to measure the length of a line segment.
Put the zero mark at point A and then move toward l to measure the length of the line segment, but it may have some errors based on the thickness of the ruler.
∗ This could be made accurate by using a Divider
Measure of an angle: Angle is formed two rays or two-line segments.
∗ In a clock, there are two hands i.e. minute hand and hour hand, which moves clockwise in every minute. When the clock hand moves from one position to another then turns through an angle.
∗To measure an angle 72^{°} using protractor-
Hence, ∠ABC = 72°.
Types of angles:
Type of angle |
Measure |
Zero angle |
0° |
Right angle |
90° |
Straight angle |
180° |
Complete angle |
360° |
Acute angle |
Between 0° to 90° |
Obtuse angle |
Between 90° to 180° |
Reflex angle |
Between 180° to 360° |
If two lines intersect with each other and form an angle of 90° then they must be perpendicular to
There several situations in our daily life, where we use these numbers to represent loss or profit; past or future; low or high temperature. The numbers on the left side of zero are called negative numbers.
Integers: The numbers which are positive, zero and negative numbers together are called as integers and they are denoted by I or Z.
Z = {…, -3, -2, -1, 0, 1, 2, 3…}.
Representation of integers on a number line: –
∴ -3 < -2 < -1 < 0 < 1 < 2 < 3 < 4 < 5 so on.
Addition and subtraction of integers:
Ex: – 3 + 2 =5, −3 – 2 = −5.
Ex: – 3 − 2 =1, −3 + 2 = −1, −10 + 5 = −5.
Addition of integers on a number line:
Add 3 and 4
∴ 3 + 4 = 7
Add −3 and −4
∴ − 3 − 4 = −7
∗ Any two distinct numbers that give zero when added to each other are additive inverse each other.
Subtraction of integers on a number line:
Subtract 3 from 6
∴ 6 − 3 = 3.
Subtract −3 from 6
On the number line, we first move 6 steps to the right of 0 to reach 6. For – 3 we have to move left but for – ( −3) we move in the opposite direction. Thus, we move 3 steps to the left of 6 and to reach 9.
∴ 6 – (−3) = 9.
A fraction means a part of a group of a whole.
The ‘whole’ here could be an object or the group of objects. But all the parts of the whole must be equal. The ‘whole’ here could be an object or the group of objects. However, all the parts of the whole must be equal.
• Fig(i) is the whole. The complete circle.
• In Fig (ii), we divide the circle into two equal parts, then the shaded portion is the half ie., of the circle.
• In Fig (iii), we divide the circle into three equal parts, then the shaded portion is the one third of the circle i.e., of the circle.
• In Fig (iv), we divide the circle into four equal parts, then the shaded portion is the one fourth of the circle i.e., of the circle.
The numerator and the denominator:
The upper part of the fraction is called ‘numerator’. It tells the no. of parts we have.
The lower part of the fraction is called ‘denominator’. It tells the total parts in whole.
Representing fractions pictorially:
Representing fractions on a number line:
Proper fractions: In a fraction if the numerator is less than denominator then, then it is called proper fraction. If we represent a proper fraction on a number line then it is always lies between 0 and 1.
Improper fractions: In a fraction if the numerator is greater than denominator then, then it is called improper fraction.
Mixed fractions: – The fraction made by the combination of whole number and a part is called mixed fraction.
Note: Only improper fractions can be represented as mixed fractions.
A mixed fraction is in the form of
We can convert it into improper fraction by
Equivalent fractions: – Equivalent fractions those fractions which represent the same part of whole.
Standard form of a fraction (simplest or lowest form):- A fraction is said to be in standard form if both the numerator and denominator of that fraction have no common factor except 1.
Like and Unlike fractions: The fractional numbers that have the same denominators are called fractional numbers and have not the same denominator are called unlike fractions.
Ex: – are like fractions and are un like fractions.
Comparing fractions:
Like fractions: – We have to compare the like fractions with the numerator only, because the like fractions have same denominator. The fraction with greater numerator is greater and the fraction with smaller numerator is smaller.
Unlike fractions: –
With same numerator: For comparing unlike fractions, we have to compare denominators when the numerator is same. The fraction with greater denominator is smaller and the fraction with smaller denominator is smaller.
Note: – To find the equivalent fractions of both the fractions with the same denominator, we have to take the LCM of their denominators.
Ascending order and Descending order: –
When we write numbers in a form that they increase from the left to right then they are in the Ascending order. When we write numbers in a form that they decrease from the left to right then they are in the Descending order.
Ex: – For fractions: are in ascending order and are in descending order.
Addition of fractions:
Like fractions: –
Un like fractions: – For adding unlike fractions, first we have to find the equivalent fraction of given fractions and then add them as like fractions.
Subtraction of fractions
Like fractions: –
Un like fractions: – First we have to find the equivalent fraction of given fractions and then subtract them as like fractions.
Decimal fractions:
A fraction where the denominator is a power of ten is called decimal fraction. We can write decimal fraction with a decimal point (.). it makes easier to do addition, subtraction and multiplication on fractions.
Data: collection of information in the form of numbers or words is called data.
Recording data: Recording of data depends on the requirement of the data. We can record data in different ways.
Organization of data: –
∗ Example for representing tally marks:
Pictograph:
If the data is represented by the picture of objects instead if numbers, then it is called pictograph. Pictures make it is easy to understand the data and answer the questions to related it by observing the pictures.
Example for representing data by pictograph
Bar graph:
• Bar graphs are used to represent the independent observations with frequencies.
• In a bar graph, bars of uniform width are drawn horizontally or vertically with equal spacing between them.
Construction of bar graph: –
Steps to construction: –
1.Draw two perpendicular lines one horizontal (x – axis) and one vertical (y – axis).
2.Along the x- axis mark ‘items’ and along the x – axis mark ‘cost of items’.
3.Select a suitable scale 1cm = 10(rupees).
4.Calculate the heights of the bars by dividing the frequencies with the scale
70 ÷ 10 = 7, 40 ÷ 10 = 4 and so on.
5.Draw rectangular vertical bars of same width on the x- axis with heights calculated above.
Algebra is the use of letters or symbols to represent number. It helps us to study about un known quantities.
Patterns:
To make a triangle, 3 matchsticks are used
For making 2 triangles we have six matchsticks
For making 3 triangles we have nine matchsticks
Thus the no. of matchsticks for making ‘n’ triangles = 3 × n = 3n.
Variable: Variable is a unknown quantity that may change. It is a dependent term.
In the above pattern, the rule is 3n, here ‘n’ is the variable.
Use of variables:
⋇ perimeter of a polygon is the sum of the lengths of all its sides.
Perimeter square = 4s, s is the variable
Perimeter of rectangle = 2 (l + b); l, b are variables.
⋇ To find the n^{th} term from the given pattern: 3, 6, 9…
Number |
3 |
6 |
9 |
12 |
15 |
… |
Pattern |
3×1 |
3×2 |
3×3 |
3×4 |
3×5 |
… |
From the table we observe that, the first number is 3×1, the second number is 3×2, the third number is 3×3 and so on.
∴ the n^{th} term of pattern 3, 6, 9, 12, = 3n, here n is variable.
Simple equation: simple equation is a condition to be satisfied by the variables. Equation has equality sign between its two sides.
Ex: 5m = 10, 2x + 1 = 0 etc.
L.H.S and R.H.S of an equation:
The expression which is at the left of equal sign of an equation is called Left Hand Side (L.H.S)
The expression which is at the right side of equal sign of an equation is called Right Hand Side (R.H.S)
Ex: 4y = 20
L.H.S = 4y and R.H.S = 20
Solution of an equation (Root of the equation):
Solution or Root of an equation is the values of variable for which L.H.S and R.H.S are equal.
Ex: 3x = 15
If x = 5; LHS = 3×5 = 15
RHS = 15
∴ solution of above equation is 5
Trial and error method:
By using this method, we get the solution of given equation.
Ex: solve 2n = 10
Substituting value of n |
Value of L. H. S |
Value of R. H. S |